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Limits of Sequences of Real Numbers

2013

Sequences of Real NumbersLimits through DefinitionsThe Squeeze Theorem

Using the Squeeze TheoremMonotonous Sequences

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VIDEO and INTERNET SUPPORT FOR THIS LECTURE

Explains the main points in THIS slide show:

http://www.youtube.com/watch?v=yBE1WApSpV4

Examples:http://www.youtube.com/watch?v=hc64LUtPjP0

Theory through examples: http://archives.math.utk.edu/visual.calculus/6/sequences.3/index.html

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Sequences of Numbers

1 2 3A ,x ,x , is a rule that assigns,

to each natural number , t

sequenc

he numb

e

.

er n

x

n x

1 1 1

1, , , ,2 4 8

1,1.4,1.41,1.414,1.4142,

Definition

Examples 1

2

1, 3,5, 7,9,3

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Limits of Sequences

1 2 3

A finite number is the of the sequence

,x ,x , if the numbers get arbitrarily close

to the number as the ind

li

ex gro

t

w

mi

.

sn

L

x x

L n

1 1 1The sequence 1, , , , converges

2 4 8

and its limit is 0.

Definition

Examples 1

If a sequence has a finite limit, then we say that the sequence is convergent or that it converges. Otherwise it diverges and is divergent.

0

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1 1 1The sequence 1, , , , converges

2 4 8

and its limit is 0.

0

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Limits of Sequences

The sequence 1,1.4,1.41,1.414,1.4142, converges

and its limit is 2.

2

3

Notation lim nnx L

The sequence (1,-2,3,-4,…) diverges.

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Computing Limits of Sequences (1)

The limit of a sequence can be often computed by inserting

in the formula defining the general term . If this expression can be

evaluated and the result is finite, then this finite value is

n

n

x n

x

the limit of

the sequence. This usually requires a rewriting of the expression . nx

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Computing Limits of Sequences (1)

1

1

1 1 1 1The limit of the sequence 1, , , , is 0 because

2 4 8 2

1inserting to the formula one gets 0.

2

n

n nn x

2 2 2

2 2

2

111 1

The limit of the sequence is 1 because rewriting 11 1 1

and inserting one gets 1.

n n nn n

nn

Examples

1

2

1n2

0

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Computing Limits of Sequences

The limit of the sequence 1 is 0 because of the rewriting n n

1 11

1

n n n nn n

n n

Insert to get the limit 0.n

Examples continued

3

1 1 .

1 1

n n

n n n n

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Formal Definition of Limits of Sequences

1 2 3

A finite number is the of the sequence

, , , if

0 : such that

lim t

.

i

n

L

x x x

n n n L x

Definition

Example1

lim 0 since if 0 is given, thenn n

1 1 10 if .n n

n n

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Visualizing the formal definition of a sequence

http://archives.math.utk.edu/visual.calculus/6/sequences.3/index.html

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Immediate consequence of the formal definition of a sequence

Every convergent sequence is bounded.

Theorem

Proof Suppose that lim xn=L . Take ϵ = 1 (any number works). Find N 1 so that whenever n > N1 we have xn within 1 of L. Then apart from the finite set { a1, a2, ... , aN} all the terms of the sequence are bounded by L+ 1 and L - 1.

So an upper bound for the sequence is max {x1 , x2 , ... , xN , L+ 1 }. Similarly one can find a lower bound.

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The Limit of a Sequence is UNIQUE

Theorem The limit of a sequence is UNIQUE

Proof Indirectly, suppose, that a sequence would have 2 limits, L1 and L2. Than for a given

∃N 1  N: n N:n>N ∈ ∀ ∈ 1 :|L1 −xn|<ϵ ∃N 2  N: n N:n>N∈ ∀ ∈  2 :| L2 −xn|<ϵ  

if N=max{N 1 ,N 2 }, xn would be arbitrary close to L1 and arbitrary close to L2 at the same, it is impossible-this is the contradiction (Unless L1 =L2)

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Calculating limit using unique prop.

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Limit of Sums

Assume that the limits lim and lim

are finite. Then lim .

n nn n

n nn

x x y y

x y x y

Let 0 be given.

To that end observe that also 0.2

Theorem

Proof

We have to find a number with the property

.n n

n

n n x y x y

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Limit of Sums

Proof1 2

1 2

Hence there are numbers and such that

and .2 2n n

n n

n n x x n n y y

1 2Let now =max , . We have

.2 2n n n n

n n n

n n x y x y x x y y

By the Triangle Inequality

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Limits of Products

2

1Let 1 and . Then lim 0 and

the limit lim does not exist. However, lim 0.

n

n n nn

n n nn n

x n y yn

x x y

The same argument as for sums can be used to prove the following result.

Assume that the limits lim and lim

are finite. Then lim .

n nn n

n nn

x x y y

x y x y

Theorem

Remark

Examples

Observe that the limits lim and lim may exist

and be finite even if the limits lim and lim do not exist.

n n n nn n

n nn n

x y x y

x y

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Squeeze Theorem for Sequences

Then the limit lim exists and

lim lim lim .

nn

n n nn n n

y

y x z

Assume that : and that

lim lim .n n n

n nn n

n x y z

x z a

Theorem

Let max , . Then

max , .

y x z

y n n n

n n n

n n a y a x a z

Let 0. Since lim lim , such

that and .

n n x zn n

x n z n

x z a n n

n n x a n n z a

Proof

This follows since .n n nx y z n

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Using the Squeeze/Pinching Theorem

Next observe that

1 2 3 1! 1 2 3 1

n

n nn n n

n n n n n n n n n n n

!Compute lim .nn

nn

Example

! 1Hence 0 .n

nn n

!Observe that 0< for all 0.n

nn

n

Solution This is difficult to compute using the standard methods because n! is defined only if n is a natural number.

So the values of the sequence in question are not given by an elementary function to which we could apply tricks like L’Hospital’s Rule.

1 !Since lim 0, also lim 0 by the Squeeze Theorem.nn n

nn n

Here each term k/n < 1.

1

.n

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Using the Squeeze Theorem

sin( )Does the sequence converge?

cos( )

If it does, find its limit.

n

n nProblem

Solution

1 1Since lim lim 0 we conclude that the sequence

-1 -1

sin( ) sin( ) converges and that lim 0.

cos( ) cos( )

n n

n

n n

n n

n n n n

1 sin( ) 1

Hence .1 cos( ) 1

n

n n n n

We have 1 sin( ) 1 and 1 cos( ) 1 for all 2,3,4, . n n n

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Monotonous SequencesDefinition

The sequence (a1,a2,a3,…) is decreasing if an+1 ≤ an for all n.

A sequence (a1,a2,a3,…) is increasing if an ≤ an+1 for all n.

The sequence (a1,a2,a3,…) is monotonous if it is either increasing or decreasing.

Theorem

The sequence (a1,a2,a3,…) is bounded if there are numbers M and m such that m ≤ an ≤ M for all n.

A bounded monotonous sequence always has a finite limit.

Observe that it suffices to show that the theorem for increasing sequences (an) since if (an) is decreasing, then consider the increasing sequence (-an).

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Monotonous SequencesTheorem A bounded monotonous sequence always has a finite limit.

Proof Let (a1,a2,a3,…) be an increasing bounded sequence.

Then the set {a1,a2,a3,…} is bounded from the above.

By the fact that the set of real numbers is complete, s=sup {a1,a2,a3,…} is finite.

lim .nna s

Claim

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Monotonous SequencesTheorem A bounded monotonous sequence always has a finite limit.

Proof Let (a1,a2,a3,…) be an increasing bounded sequence.

Let s=sup {a1,a2,a3,…}.

lim .nna s

Claim

Proof of the Claim Let 0.

We have to find a number with the property that .nn n n a s

.Since sup , there is an element such that n n n ss a a s a

Since is increasing .n n na n n s a a s

Hence .nn n a s This means that lim .nna s

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SUMMARY

1. Notion of a sequence 2. Notion of a limit of a sequence 3. The limit of a convergent sequence is

unique. 4. Every convergent sequence is bounded.

5. Any bounded increasing (or decreasing) sequence is convergent.Note that if the sequence is increasing (resp. decreasing), then the limit is the least-upper bound (resp. greatest-lower bound) of the numbers

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SUMMARY

6. If two sequences are convergent and we compose their +, -, *. /, 1/.. then the limit of this composed sequence exists and is the +, -, *. /, 1/..of the original limiting values.

7. Squeeze/Pinching theorem