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11INFINITE SEQUENCES AND SERIES
Infinite sequences and series were introduced briefly in A Preview of Calculusin connection with Zenorsquos paradoxes and the decimal representation of numbers
INFINITE SEQUENCES AND SERIES
Their importance in calculus stems from Newtonrsquos idea of representing functions as sums of infinite series
For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series
INFINITE SEQUENCES AND SERIES
We will pursue his idea in Section 1110 in order to integrate such functions as e-x2
Recall that we have previously been unable to do this
INFINITE SEQUENCES AND SERIES
Many of the functions that arise in mathematical physics and chemistry such as Bessel functions are defined as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another way as we will see in Section 1111
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
111Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
Infinite sequences and series were introduced briefly in A Preview of Calculusin connection with Zenorsquos paradoxes and the decimal representation of numbers
INFINITE SEQUENCES AND SERIES
Their importance in calculus stems from Newtonrsquos idea of representing functions as sums of infinite series
For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series
INFINITE SEQUENCES AND SERIES
We will pursue his idea in Section 1110 in order to integrate such functions as e-x2
Recall that we have previously been unable to do this
INFINITE SEQUENCES AND SERIES
Many of the functions that arise in mathematical physics and chemistry such as Bessel functions are defined as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another way as we will see in Section 1111
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
111Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
Their importance in calculus stems from Newtonrsquos idea of representing functions as sums of infinite series
For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series
INFINITE SEQUENCES AND SERIES
We will pursue his idea in Section 1110 in order to integrate such functions as e-x2
Recall that we have previously been unable to do this
INFINITE SEQUENCES AND SERIES
Many of the functions that arise in mathematical physics and chemistry such as Bessel functions are defined as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another way as we will see in Section 1111
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
111Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
We will pursue his idea in Section 1110 in order to integrate such functions as e-x2
Recall that we have previously been unable to do this
INFINITE SEQUENCES AND SERIES
Many of the functions that arise in mathematical physics and chemistry such as Bessel functions are defined as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another way as we will see in Section 1111
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
111Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
Many of the functions that arise in mathematical physics and chemistry such as Bessel functions are defined as sums of series
It is important to be familiar with the basic concepts of convergence of infinite sequences and series
INFINITE SEQUENCES AND SERIES
Physicists also use series in another way as we will see in Section 1111
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
111Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
Physicists also use series in another way as we will see in Section 1111
In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it
INFINITE SEQUENCES AND SERIES
111Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
111Sequences
In this section we will learn about
Various concepts related to sequences
INFINITE SEQUENCES AND SERIES
SEQUENCE
A sequence can be thought of as a list of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCE
A sequence can be thought of as a list of numbers written in a definite order
a1 a2 a3 a4 hellip an hellip
The number a1 is called the first term a2 is the second term and in general an is the nth term
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
We will deal exclusively with infinite sequences
So each term an will have a successor an+1
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
Notice that for every positive integer n there is a corresponding number an
So a sequence can be defined as
A function whose domain is the set of positive integers
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
However we usually write an instead of the function notation f(n) for the value of the function at the number n
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
The sequence a1 a2 a3 is also denoted by
1orn n n
a a infin
=
Notation
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
Some sequences can be defined by giving a formula for the nth term
Example 1
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
In the following examples we give three descriptions of the sequence
1 Using the preceding notation
2 Using the defining formula
3 Writing out the terms of the sequence
Example 1
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
In this and the subsequent examples notice that n doesnrsquot have to start at 1
Example 1 a
Preceding Notation
Defining Formula
Terms of Sequence
11 n
nn
infin
=
+ 1n
nan
=+
1 2 3 4 2 3 4 5 1
nn
+
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES Example 1 b
Preceding Notation
Defining Formula
Terms of Sequence
( 1) ( 1)3
n
n
n minus +
( 1) ( 1)3
n
n n
na minus +=
2 3 4 5 3 9 27 81
( 1) ( 1) 3
n
n
n
minus minus minus +
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES Example 1 c
Preceding Notation
Defining Formula
Terms of Sequence
3
3n
ninfin
=minus 3
( 3)na n
n= minus
ge 01 2 3 3n minus
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES Example 1 d
Preceding Notation
Defining Formula
Terms of Sequence
0
cos6 n
nπ infin
=
cos6
( 0)
nna
n
π=
ge
3 11 0 cos 2 2 6
nπ
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
Find a formula for the general term an
of the sequence
assuming the pattern of the first few terms continues
3 4 5 6 7 5 25 125 625 3125
minus minus
Example 2
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
We are given that
1 2 3
4 5
3 4 55 25 125
6 7625 3125
a a a
a a
= = minus =
= minus =
Example 2
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term
The second term has numerator 4 and the third term has numerator 5
In general the nth term will have numerator n+2
Example 2
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
The denominators are the powers of 5
Thus an has denominator 5n
Example 2
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
The signs of the terms are alternately positive and negative
Hence we need to multiply by a power of ndash1
Example 2
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
In Example 1 b the factor (ndash1)n meant we started with a negative term
Here we want to start with a positive term
Example 2
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
Thus we use (ndash1)nndash1 or (ndash1)n+1
Therefore
1 2( 1)5
nn n
na minus += minus
Example 2
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
We now look at some sequences that donrsquot have a simple defining equation
Example 3
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
The sequence pn where pn
is the population of the world as of January 1 in the year n
Example 3 a
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
If we let an be the digit in the nth decimal place of the number e then an is a well-defined sequence whose first few terms are
7 1 8 2 8 1 8 2 8 4 5 hellip
Example 3 b
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
FIBONACCI SEQUENCE
The Fibonacci sequence fn is defined recursively by the conditions
f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3
Each is the sum of the two preceding term terms
The first few terms are 1 1 2 3 5 8 13 21 hellip
Example 3 c
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
FIBONACCI SEQUENCE
This sequence arose when the 13th-century Italian mathematician Fibonacci solved a problem concerning the breeding of rabbits
See Exercise 71
Example 3
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
A sequence such as that in Example 1 a [an = n(n + 1)] can be pictured either by Plotting its terms on
a number line Plotting its graph
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
Note that since a sequence is a function whose domain is the set of positive integers its graph consists of isolated points with coordinates
(1 a1) (2 a2) (3 a3) hellip (n an) hellip
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
From either figure it appears that the terms of the sequence an = n(n + 1) are approaching 1 as nbecomes large
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
In fact the difference
can be made as small as we like by taking nsufficiently large
We indicate this by writing
111 1
nn n
minus =+ +
lim 11n
nnrarrinfin
=+
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
In general the notation
means that the terms of the sequence anapproach L as n becomes large
lim nna L
rarrinfin=
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SEQUENCES
Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 26
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if we can make the terms an as close to Las we like by taking n sufficiently large
If exists the sequence converges(or is convergent)
Otherwise it diverges (or is divergent)
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 1
lim nna
rarrinfin
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
Here Definition 1 is illustrated by showing the graphs of two sequences that have the limit L
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
A more precise version of Definition 1 is as follows
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
A sequence an has the limit L and we write
if for every ε gt 0 there is a corresponding integer N such that
if n gt N then |an ndash L| lt ε
lim or asn nna L a L n
rarrinfin= rarr rarrinfin
Definition 2
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
Definition 2 is illustrated by the figure in which the terms a1 a2 a3 are plotted on a number line
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
No matter how small an interval (L ndash ε L + ε) is chosen there exists an N such that all terms of the sequence from aN+1 onward must lie in that interval
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
Another illustration of Definition 2 is given here
The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
This picture must be valid no matter how small ε is chosen
Usually however a smaller ε requires a larger N
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
If you compare Definition 2 with Definition 7 in Section 26 you will see that the only difference between and is that n is required to be an integer
Thus we have the following theorem
lim nna L
rarrinfin= lim ( )
xf x L
rarrinfin=
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
If and f(n) = an when nis an integer then
lim ( )x
f x Lrarrinfin
=
lim nna L
rarrinfin=
Theorem 3
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Theorem 3 is illustrated here
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
In particular since we know that when r gt 0 (Theorem 5
in Section 26) we havelim(1 ) 0r
xx
rarrinfin=
Equation 4
1lim 0 if 0rnr
nrarrinfin= gt
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
If an becomes large as n becomes large we use the notation
The following precise definition is similar to Definition 9 in Section 26
lim nna
rarrinfin= infin
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT OF A SEQUENCE
means that for every positive number M there is an integer N such that
if n gt N then an gt M
Definition 5
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
If then the sequence an is divergent but in a special way
We say that an diverges to infin
lim nna
rarrinfin= infin
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
The Limit Laws given in Section 23 also hold for the limits of sequences and their proofs are similar
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT LAWS FOR SEQUENCES
Suppose an and bn are convergent sequences and c is a constant
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT LAWS FOR SEQUENCES
Then
lim( ) lim lim
lim( ) lim lim
lim lim lim
n n n nn n n
n n n nn n n
n nn n n
a b a b
a b a b
ca c a c c
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
rarrinfin rarrinfin rarrinfin
+ = +
minus = minus
= =
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMIT LAWS FOR SEQUENCES
Alsolim( ) lim lim
limlim if lim 0
lim
lim lim if 0 and 0
n n n nn n n
nn nnn n
n nn
PPn n nn n
a b a b
aa bb b
a a p a
rarrinfin rarrinfin rarrinfin
rarrinfin
rarrinfin rarrinfinrarrinfin
rarrinfin rarrinfin
= sdot
= ne
= gt gt
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
SQUEEZE THEOREM FOR SEQUENCES
If an le bn le cn for n ge n0
and then
lim nnb L
rarrinfin=
lim limn nn na c L
rarrinfin rarrinfin= =
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Another useful fact about limits of sequences is given by the following theorem
The proof is left as Exercise 75
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES Theorem 6
lim 0nna
rarrinfin=
lim 0nna
rarrinfin=
If
then
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Find
The method is similar to the one we used in Section 26
We divide the numerator and denominator by the highest power of n and then use the Limit Laws
Example 4
lim1n
nnrarrinfin +
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Thus
Here we used Equation 4 with r = 1
lim11lim lim 1 11 1 lim1 lim
1 11 0
n
n n
n n
nn
n n
rarrinfin
rarrinfin rarrinfin
rarrinfin rarrinfin
= =+ + +
= =+
Example 4
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Calculate
Notice that both the numerator and denominator approach infinity as n rarr infin
Example 5
lnlimn
nnrarrinfin
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Here we canrsquot apply lrsquoHospitalrsquos Rule directly
It applies not to sequences but to functions of a real variable
Example 5
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
However we can apply lrsquoHospitalrsquos Rule to the related function f(x) = (ln x)x and obtain
Example 5
ln 1lim lim 01x x
x xxrarrinfin rarrinfin
= =
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Therefore by Theorem 3 we have
Example 5
lnlim 0n
nnrarrinfin
=
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Determine whether the sequence an = (ndash1)n
is convergent or divergent
Example 6
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
If we write out the terms of the sequence we obtain
ndash1 1 ndash1 1 ndash1 1 ndash1 hellip
Example 6
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
The graph of the sequence is shown
The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number
Example 6
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Thus does not exist
That is the sequence (ndash1)n is divergent
Example 6
lim( 1)n
nrarrinfinminus
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Evaluate if it exists
Thus by Theorem 6
Example 7( 1)lim
n
n nrarrinfin
minus
( 1) 1lim lim 0n
n nn nrarrinfin rarrinfin
minus= =
( 1)lim 0n
n nrarrinfin
minus=
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
The following theorem says that if we apply a continuous function to the terms of a convergent sequence the result is also convergent
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
If and the function f is continuous at L then
The proof is left as Exercise 76
Theorem 7
lim nna L
rarrinfin=
lim ( ) ( )nnf a f L
rarrinfin=
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Find
The sine function is continuous at 0 Thus Theorem 7 enables us to write
Example 8
lim sin( )n
nπrarrinfin
( )lim sin( ) sin lim( )
sin 0 0n n
n nπ πrarrinfin rarrinfin
=
= =
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Discuss the convergence of the sequence an = nnn where n = 1 2 3 hellip n
Example 9
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Both the numerator and denominator approach infinity as n rarr infin
However here we have no corresponding function for use with lrsquoHospitalrsquos Rule
x is not defined when x is not an integer
Example 9
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Letrsquos write out a few terms to get a feeling for what happens to an as n gets large
Example 9
1 2 31 2 1 2 312 2 3 3 3
a a asdot sdot sdot= = =
sdot sdot sdot
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Therefore
1 2 3n
nan n n nsdot sdot sdot sdot sdot sdot sdot
=sdot sdot sdot sdot sdot sdot sdot
E g 9mdashEquation 8
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
From these expressions and the graph here it appears that the terms are decreasing and perhaps approach 0
Example 9
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
To confirm this observe from Equation 8 that
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator
Example 9
1 2 3n
nan n n n
sdot sdot sdot sdot sdot sdot = sdot sdotsdot sdot sdot sdot
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Thus 0 lt an le
We know that 1n rarr 0 as n rarr infin
Therefore an rarr 0 as n rarr infin by the Squeeze Theorem
Example 91n
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
For what values of r is the sequence r n convergent
From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1
Example 10
lim x
xa
rarrinfin= infin
lim 0x
xa
rarrinfin=
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
Thus putting a = r and using Theorem 3 we have
It is obvious that
Example 10
if 1lim
0 if 0 1n
n
rr
rrarrinfin
infin gt= lt lt
lim1 1 and lim 0 0n n
n nrarrinfin rarrinfin= =
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
If ndash1 lt r lt 0 then 0 lt |r | lt 1
Thus
Therefore by Theorem 6
Example 10
lim | | lim | | 0n n
n nr r
rarrinfin rarrinfin= =
lim 0n
nr
rarrinfin=
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
If r le ndash1 then r n diverges as in Example 6
Example 10
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
The figure shows the graphs for various values of r
Example 10
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
The case r = ndash1 was shown earlierExample 10
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
The results of Example 10 are summarized for future use as follows
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
The sequence r n is convergent if ndash1 lt r le 1 and divergent for all other values of r
Equation 9
0 if 1 1lim
1 if 1n
n
rr
rrarrinfin
minus lt lt= =
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
LIMITS OF SEQUENCES
A sequence an is called
Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt
Decreasing if an gt an+1 for all n ge 1
Monotonic if it is either increasing or decreasing
Definition 10
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
DECREASING SEQUENCES
The sequence is decreasing because
and so an gt an+1 for all n ge 1
Example 113
5n +
3 3 35 ( 1) 5 6n n ngt =
+ + + +
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
DECREASING SEQUENCES
Show that the sequence
is decreasing
Example 12
2 1nna
n=
+
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
DECREASING SEQUENCES
We must show that an+1 lt an that is
2 2
1( 1) 1 1
n nn n
+lt
+ + +
E g 12mdashSolution 1
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
DECREASING SEQUENCES
This inequality is equivalent to the one we get by cross-multiplication
2 22 2
3 2 3 2
2
1 ( 1)( 1) [( 1) 1]( 1) 1 1
1 2 21
n n n n n nn n
n n n n n nn n
+lt hArr + + lt + +
+ + +
hArr + + + lt + +
hArr lt +
E g 12mdashSolution 1
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
DECREASING SEQUENCES
Since n ge 1 we know that the inequality n2 + n gt 1 is true
Therefore an+1 lt an
Hence an is decreasing
E g 12mdashSolution 1
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
DECREASING SEQUENCES
Consider the function2( )
1xf x
x=
+
2 2
2 2
22
2 2
1 2( )( 1)1 0 whenever 1
( 1)
x xf xx
x xx
+ minus=
+
minus= lt gt
+
E g 12mdashSolution 2
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
DECREASING SEQUENCES
Thus f is decreasing on (1 infin)
Hence f(n) gt f(n + 1)
Therefore an is decreasing
E g 12mdashSolution 2
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
A sequence an is bounded
Above if there is a number M such that an le Mfor all n ge 1
Below if there is a number m such that m le anfor all n ge 1
If it is bounded above and below
Definition 11
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
For instance
The sequence an = n is bounded below (an gt 0) but not above
The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
We know that not every bounded sequence is convergent
For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
Similarly not every monotonic sequence is convergent (an = n rarr infin)
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
However if a sequence is both bounded and monotonic then it must be convergent
This fact is proved as Theorem 12
However intuitively you can understand why it is true by looking at the following figure
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
If an is increasing and an le M for all n then the terms are forced to crowd together and approach some number L
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
The proof of Theorem 12 is based on the Completeness Axiom for the set of real numbers
This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
This means
b is an upper bound for S
However if M is any other upper bound then b le M
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
BOUNDED SEQUENCES
The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
Every bounded monotonic sequence is convergent
Theorem 12MONOTONIC SEQ THEOREM
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
Suppose an is an increasing sequence
Since an is bounded the set S = an|n ge 1 has an upper bound
By the Completeness Axiom it has a least upper bound L
Theorem 12mdashProof
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
Given ε gt 0 L ndash ε is not an upper bound for S (since L is the least upper bound)
Therefore aN gt L ndash ε for some integer N
MONOTONIC SEQ THEOREM Theorem 12mdashProof
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
However the sequence is increasing
So an ge aN for every n gt N
Thus if n gt N we have an gt L ndash ε
Since an le L thus 0 le L ndash an lt ε
MONOTONIC SEQ THEOREM Theorem 12mdashProof
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
Thus |L ndash an| lt ε whenever n gt N
Therefore
Theorem 12mdashProof
lim nna L
rarrinfin=
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
A similar proof (using the greatest lower bound) works if an is decreasing
Theorem 12mdashProof
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent
Likewise a decreasing sequence that is bounded below is convergent
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
This fact is used many times in dealing with infinite series
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
Investigate the sequence an defined by the recurrence relation
Example 13
11 1 22 ( 6) for 1 23n na a a n+= = + =
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
We begin by computing the first several terms
These initial terms suggest the sequence is increasing and the terms are approaching 6
Example 13
1 11 2 32 2
14 5 62
7 8 9
2 (2 6) 4 (4 6) 5(5 6) 55 575 5875
59375 596875 5984375
a a aa a aa a a
= = + = = + =
= + = = =
= = =
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
To confirm that the sequence is increasing we use mathematical induction to show that an+1 gt an for all n ge 1
Mathematical induction is often used in dealing with recursive sequences
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
That is true for n = 1 because a2 = 4 gt a1
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
If we assume that it is true for n = kwe have
Hence
and
Thus
Example 13
1k ka a+ gt
1 6 6k ka a+ + gt +1 1
12 2( 6) ( 6)k ka a+ + gt +
2 1k ka a+ +gt
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
We have deduced that an+1 gt an is true for n = k + 1
Therefore the inequality is true for all nby induction
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
Next we verify that an is bounded by showing that an lt 6 for all n
Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
We know that a1 lt 6
So the assertion is true for n = 1
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
Suppose it is true for n = k
Then
Thus
and
Hence
6ka lt
6 12ka + lt
Example 13
1 12 2( 6) (12) 6ka + lt =
1 6ka + lt
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
This shows by mathematical induction that an lt 6 for all n
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
Since the sequence an is increasing and bounded Theorem 12 guarantees that it has a limit
However the theorem doesnrsquot tell us what the value of the limit is
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
Nevertheless now that we know exists we can use the recurrence relation to write
lim nnL a
rarrinfin=
11 2
12
12
lim lim ( 6)
(lim 6)
( 6)
n nn n
nn
a a
a
L
+rarrinfin rarrinfin
rarrinfin
= +
= +
= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
MONOTONIC SEQ THEOREM
Since an rarr L it follows that an+1 rarr L too (as n rarr infin n + 1 rarr infin too)
Thus we have
Solving this equation for L we get L = 6 as predicted
12 ( 6)L L= +
Example 13
