1

Chapter 9: Infinite Series

In this Chapter we will be studying infinite series, which is just another name for infinite sums. You have

studied some of these in the past when you looked at infinite geometric sums of the form:

2 31 1 1

12 2 2

. Does it ring a bell?

In this Chapter, we consider a more general point of view, where the main question will be to determine which

infinite sums add up to something finite and which do not (which series converge and which not, similar with

the definitions for improper integrals). To answer to this question, we will develop a series of tests, the

collection of which should cover many types of series. In this sense, studying series is like studying integrals,

we use certain types of tests for certain series, and an important part of solving a problem is to determine which

test to use for which series.

Since any series is a sequence, we will begin the chapter with reviewing sequences.

TABLE OF CONTENTS:

9.1. Infinite sequences

9.2. Introduction to infinite series

9.3. Positive series: The integral Test

9.4. Positive series: Other Tests

9.5. Alternating series. Absolute convergence and conditional convergence

9.6. Power Series

9.7. Operations on Power Series

9.8. Taylor and MacLaurin Series

9.9. The Taylor Approximation Theorem

2

9.1. Infinite sequences:

Definition: A sequence 1,{ }n na is an ordered (indexed) list of numbers.

Ways to define a sequence:

list the terms of a sequence. Example: (1) 1,4,9,14, ;

give a formula (definition) for the general term of the sequence: (2) 5( 1) 1 5 6, 1na n n n ;

This is the preferred method since it is so easy to use (plot, calculate limit, etc.) and we often need to

obtain a formula like (2) from the other forms ((1) or (3));

give a recurrence relation and the initial term: (3) 1 11; 5, 2n na a a n ; or the Fibonacci

sequence: 1 2 1 21; 1; , 3n n nf f f f f n

Examples of sequences:

1 , n 1n

na 2 , n 1na n sin( ), n 2na n

Note that n doesn’t need to start at 1 in the definition of a sequence.

NOTE: Note that any sequence can be viewed as a particular function :f N R , ( ) nf n a . Therefore,

when plotting a sequence, we plot na versus n , as plotting a function.

The most important question for a sequence is its convergence.

Definition: A sequence converges to a real number L , and we denote this by lim nn

a L

if

0, such that , nN N a L n N .

(This is the definition for the convergence of a function at , that is the terms of the sequence get as close as

we want to L if n is big enough).

A sequence which is not convergent is divergent.

Examples: 1 , n 1n

na ; 2 , n 1na n ; sin( ), n 1na n ; 1 , n 1n

na n ;

1

, n 1

n

nan

;

1, n 1n

na

n

-- Plot terms to guess L and prove L by definition.

3

More efficient methods to calculate limits are discussed below.

NOTE: If lim nn

a

, we say that na diverges to . (Keep this is mind) .

NOTE: A sequence is convergent if it has a finite limit at infinity.

NOTE: Since any sequence is a function, when studying the limit of a sequence we really study the limit of a

function at . Therefore, all previous tools which we have for calculating the limit of a function (limits of

elementary functions, operations with limits, fundamental limits, and a special form of l’Hospital’s Rule (which

will be presented soon) can be used).

To calculate the limit of a sequence we’ll use:

--- limits of elementary functions (polynomials, rationals, exponential, etc) / expressions, operations with

limits;

--- fundamental limits, i.e. 1

0

, 1

0 ;

--- adapted form of l’Hospital rule – see below;

--- squeeze theorem;

--- monotonic sequence theorem .

Let us develop each of these.

The following rules are true:

Theorem 1: If na and nb are convergent sequences and k is a (real) constant, then:

lim

lim lim

lim lim lim (1)

lim lim lim

limlim , lim 0

lim

n

n nn n

n n n nn n n

n n n nn n n

nn n

nn n

n nn

k k

ka k a

a b a b

a b a b

aaif b

b b

Examples: Calculate:

a) 2

2 1lim

3 5n

n

n

b) lim

1n

n

n c)

2lim

ln( )n

n

n

4

In the above example c), we cannot apply l’Hospital’s Rule directly (cannot take the derivative of a sequence –

discrete set of points --) . Instead, we use the following:

Theorem 2:

If lim ( )x

f x L

then lim ( ) lim nn n

f n a L

.

Proof: The result is obvious (if a continuous function converges, then the sequence which represents the

function discretely converges to the same limit). It can be proved by definition.

Therefore, when we need to use l’Hospital’s Rule in a sequence, ALWAYS transform it first to a function:

'

.2

2 2 2lim lim lim lim 2

1ln( ) ln( )

l H

n Th x x x

n xx

n x

x

.

Never take the derivative of a sequence, but apply the method above (otherwise points will be deducted ).

Another useful result to calculate limits of sequences:

Theorem 3 (squeeze theorem) : Let na and nc be convergent sequences to the same limit L such that

lim nn

a L

and lim nn

c L

and let

(2) , n n na b c n K , where K is an arbitrary natural number . Then lim nn

b L

.

The theorem above is represented graphically :

The squeeze theorem is used often with trigonometric functions sin( )n and cos( )n which can be bounded, but

cannot be evaluated directly at .

Example: sin( )

limn

n

n Plot na , nb and nc on your calculator in this case.

5

Theorem 4 (monotonic sequence theorem):

If na is an increasing sequence ( 1, n na a n K ) and bounded above

( such that , nU a U n K ), then na converges to L U .

Similarly, If na is a decreasing sequence ( 1, n na a n K ) and bounded below

( such that , nB a B n K ), then na converges to L B .

Intuitively this theorem should make sense. In the first case, since the sequence is increasing and not going to

, it should approach a finite limit. The proof is based on showing that na converges to the least upper

bound of , 1nS a n using the properties that L is an upper bound, but L is not.

Examples:

Determine if the following sequences are convergent, and if they are, calculate lim nn

a

:

a) 1 1 1

12! 3! !

nan

Hint: Show that na is increasing and that

1 1 1 11 2 2

1 2 2 3 ( 1)na

n n n

(Problem 34)

b) 2

2n n

nb Show that 1n nb b by forming perfect square and 0nb .

c) 1 1

11; 1 , 1

2n na a a n . List a few terms to guess monotonicity, show that na is increasing if

bounded above (which we also show, that is 2na ).

d) 1 1

12; 6 , 1

2n na a a n . List a few terms to guess monotonicity, show that na is increasing if

bounded above (which we also show, that is 6na ).

These methods (monotonic sequence theorem and squeeze theorem) are the only feasible methods to use for

these types of problems (recurrent sequences and trigonometric sequences) so learn them well.

Homework:

What left from above plus problems 1,2,4,6,9,17,23,25,31,33,36,37,45. E.C. 50, 53

E.C.: Find the limit of: 2, 2 2 , 2 2 2 ,

6

9.2. Introduction to infinite series:

Definition: An infinite series (singular same as plural) is an infinite sum.

Given a sequence 1n n

a

, the infinite series 1 2 na a a is denoted by

1

n

n

a

.

We have studied some finite sums before:

( 1)1 2

2

n nn

,

11

1 1 1 21

12 4 21

2

n

n

, et. al, but now we want to look at infinite sums.

For example, what would 1 2 n be ? What about 1 1 1

12 4 2

n

Infinite series have many areas of application:

Decimal representation of numbers;

Example: 1

3

1 3 3 3 1100.3 0.333

13 10 100 10 31

10

n

n

does it work out?

Do the same for 1

9,

2

3.

The definition of the definite integral: 1

1

( ) lim

b n

k k kn

ka

f x dx f x x

Representation of analytical functions as power series : in your calculator most functions which are not

polynomial (e.g. x , xe , ln( )x ) are represented first as infinite polynomials (power series) before they

are evaluated. We will see how this happens in the final sections of this chapter.

Solution of some differential equations: analytical series method for linear or nonlinear differential

equations (Cauchy) or general theory for non-linear differential equations around critical points. …

Like we saw above, some series “diverge” (to ∞), while others may converge to a finite value. This issue of

convergence for a series is the most important in the study of series and central to this chapter.

7

In order to study the convergence of an infinite series1

n

n

a

, it is most convenient to consider the partial sums

1 2

1

n

n k n

k

S a a a a

. Note that the partial sums form a sequence in themselves 1n n

S

. To see if a

sequence converge, all we have to do is to calculate lim nn

s

.

Definition: The infinite series 1

n

n

a

is called convergent if the sequence of partial sums 1n n

S

converges (to

a real value), that is if lim nn

S S

. Otherwise (if lim nn

S

or if lim nn

S

does not exist), the series

1

n

n

a

is divergent.

Notice that 1

n

n

a

=1

lim limn

n kn n

k

S a

. For convergent series, lim nn

S S

is called the sum of the series.

Examples:

Is 1n

n

convergent? What about 1

1

2

n

n

?

In general, an infinite geometric series is a series of the form

2 1 1

1

n n

n

a a r a r a r a r

We know that

2 1 1

1

1

1

nn

n k

n

k

a rS a a r a r a r a r

r

.

Therefore, , if 1

lim 1

or D.N.E. if 1n

n

ar

S r

r

.

This is a general result that we should memorize: An infinite geometric series of the form

2 na a r a r a r is convergent to 1

a

rwhen the absolute ratio 1r , and is divergent

otherwise.

Examples: conclude directly about 1

1

2

n

n

, 4 4 4 1

43 9 27 3

n

,

4 16 64 4

3 9 27 3

n

8

1

110 20 40 25 5 1

3 9 27 3

nn

, 2

1

2 3n n

n

Telescopic series: They are series (usually formed from fractions) in which the terms cancel

consecutively if written appropriately.

Example: 1

1

( 1)n n n

,

1 1

1

( 1) 1

n n

n

k k

A BS

k k k k

, we find that 1A and 1B such that

1

1 1 1 1 1 1 1 1 1 1 1 11 1

1 2 2 3 3 4 1 1 1

n

n

k

Sk k n n n n n

, and

therefore

1

1

( 1)n n n

is convergent to 1. For simple fractions like above, it is worthwhile to see if it is a

telescoping series.

Theorem 1 (The Divergence Test --- n’th Term Divergence Test):

If a general series 1

n

n

a

is convergent, then lim 0nn

a

.

Conversely, if lim 0nn

a

, then 1

n

n

a

is divergent.

NOTE: This test (the first general test that we have) is very important and it will help us identify many

divergent series. However, keep in mind that this test is useful to recognize divergent series, and not convergent

series (it is conclusive only when 1

lim 0 ( )n nn

n

a a D

, but inconclusive when lim 0nn

a

), i.e. the

condition lim 0nn

a

is only a NECESSARY condition for convergence, and NOT A SUFFICIENT

condition.

Examples: Try to apply the divergence test (or other tools) to study the convergence of the following series:

1n

n

, 1 1n

n

n

,

1 ln( )n

n

n

, 1 ln( )n

n

n

, 1

1

n n

, 2

1

1

n n

For the last two series, the divergence test is inconclusive. Actually, we can show that 1

1

n n

( C) and that 2

1

1

n n

(D).

For 1

1

n n

, show that 2

12

n

nS by considering 2 4 8, , , .S S S etc .

21

1

n n

will be studied a bit later.

9

Theorem 2 (operations with series):

If 1

n

n

a

and 1

n

n

b

are convergent series, then so are the series 1

n

n

ca

, 1

n n

n

a b

and 1

n n

n

a b

and

1 1

n n

n n

ca c a

, 1 1 1

n n n n

n n n

a b a b

, 1 1 1

n n n n

n n n

a b a b

.

NOTE: The above theorem doesn’t say anything about the case when one of the series (or both) are divergent.

The case 1

n

n

a

( C) and 1

n

n

b

(D) usually implies that 1

n n

n

a b

(D) . Why? However, the case 1

n

n

a

(D)

and 1

n

n

b

(D) is harder.

Examples: 1

3 1

( 1) 2nn n n

,

1

3 1

( 1)n n n n

,

1

1 1

1n n n

.

Homework:

What left from above, plus 1,2,5,6,7,9,12,18,21,23,26. E.C.: 29,34,50

10

9.3. Positive series: The integral Test:

The focus of the next sections is to determine the convergence of a series, although occasionally we will be

able to determine estimates for the sum of some series too. We will derive a series of tests (of the type of the

divergence test presented in 9.2) which we’ll help us determine immediately if a series is convergent or

divergent.

For the next two sections, we restrict to the study of positive series only (series for which most terms are

positive i.e. 1

n

n

a

with 0, for na n K ). This is because it is much easier to devise convergence tests for

positive series. However, do not forget this restriction, and always check that the series is a positive series

before applying any of these tests!

Example: Which of the following are positive series?

1n

n

, 1

1

n n

, 1

sin( )

n

n

n

,

1

1n

n n

,

1

1

n n

, 1

cos( )

n

n

n

The only danger of divergence for positive series is that they can add up to , since the partial sums are always

increasing. You can read the Bounded Sum Test which discusses that.

Optional:

Lemma (Bounded Sum Test):

A positive series 1

n

n

a

is convergent if and only the sequence of its partial sums 1

n

n k

k

S a

is bounded from

above.

Proof: ←: Note that nS is increasing, since 1 1 0n n nS S a (positive series). If nS is upper bounded,

then due to the monotonic sequence theorem, it is convergent.

→: If 1

n

n

a

is convergent, then nS is convergent to a finite value S . Using the definition of convergence,

this implies that nS is bounded.

NOTE: The result above is a Lemma, and we will use it independently only rarely. However, it is paramount

for positive series and is used in the following theorem.

Example: 1

1

!n n

End Optional

11

Theorem (The Integral Test for Positive Series):

Consider the positive series 1

n

n

a

and the corresponding function :[1, ) [0, )f , ( ) nf n a , which is

assumed to be a continuous, positive and decreasing function ( '( ) 0, [1, )f x x ). Then the infinite series

1

n

n

a

converges if and only if the improper integral 1

( )f x dx

converges.

NOTE: This is a very important result, which reduces the study of the convergence of a series to the

convergence of an improper integral (much easier to evaluate). However, before using it, you SHOULD

ALWAYS CHECK THAT the function f(x) is continuous, positive and especially DECREASING over its

domain.

NOTE: The integral test is used when all conditions above are met (and relatively easy to check), and when we

are able to calculate relatively easily 1

( )f x dx

.

NOTE: The integral test is a necessary and sufficient condition (it goes both ways): therefore, it can be used for

both convergent and divergent series.

Examples: Use the integral test above (or other test) to determine the convergence of the following series:

1

2n

n

, 1

arctan( )n

n

,

0

1

2n

n

,

1

1n

n n

,

1

1

n n

, 2

1

1

n n

, 2

1

sin( )

n

n

n

, 1

1

!n n

, 1

sin2n

n

Let us look in general at the p-series: 1

1p

n n

, 0p for now.

The function :[1, ) [0, )f , 1

( )p

f xx

is positive and continuous. Also,

1

1'( ) 0p

p

pf x px

x

over the function’s domain, so ( )f x is decreasing. All conditions of the

integral test are met.

12

1

1 1

1, when 1 0 1

1

, when 11

, when 0< 1

pp

p pp

xx dx p

pp

. Therefore, 1

( ) for 11 is

( ) for 1pn

C p

D pn

.

(you could also see directly that 1

1p

n n

(D) for 0p -- why?) .

This is an important result for series which we should memorize. Combined with some other tests/next section,

we should be able to determine the convergence of many new series, and especially of almost any series

produced by rational and irrational functions!

Examples: Determine the convergence of the following series:

1

1

n n n

, 100

1

1

n n

, 1

1

n n

, 1n

n

, 1

1

n n

, 1

1

1n n

,

11 3

n

nn

e

, 3

5

2n n

,

41 1n

n

n

,

1

1

ln( )n n n

, 2

1

n

n

ne

,

21

1

4n n

1. The Riemann zeta-function is defined by 1

1( )

xn

xn

and is used in number theory to study the

distribution of prime numbers. What is the domain of ?

Proof (of the integral test):

Figure 9.2.1 The function ( ) nf n a and the sequence na

13

Consider a decreasing positive continuous function :[1, ) [0, )f , given by ( ) nf n a .

Looking at a picture above, it is easy to see that (1) 2 3 1 3 1

1

( )

n

n na a a f x dx a a a

Therefore, (2)

1

2 11

( )

nn n

k k

k k

a f x dx a

. If 1

( )f x dx

is convergent, then 1

( )

n

f x dx U (this sequence is

convergent, and therefore it is upper bounded, since increasing), so from the inequality (2) we have that

1

1

n

n k

k

S a a U

(upper bounded). Since 1 1 0n n nS S a ( nS is increasing), from the monotonic

sequence theorem, it follows that nS is convergent. Q.E.D. Conversely, if 1

n

n

a

( C) → nS → nS is upper

bounded (since increasing) , Therefore, from (2) it follows that the sequence 1

( )

n

f x dx is upper bounded (and

increasing), so by the monotonic sequence theorem, 1

( )

n

f x dx (and therefore 1

( )f x dx

) is convergent. Q.E.D.

∎

An estimate for the sum of a series:

The tests before only determine if a series is convergent or divergent, but how can we determine (or at least

estimate) the sum of a series. For a positive series, and when the conditions of the integral test are met, we can

find an estimate for nS S , which represents the error that we make by approximating the sum of the series

with its partial sum nS :

14

Figure 9.2.2 Estimate of the error nS S using the integral test

From Figure 9.2.2., we see that:

1 2 ( )n n n n

n

S S S S a a f x dx

, which gives us a very useful estimate for the error nS S .

Example: Give an estimate of the error made by approximating the sum of the series 2

1

1

n n

with taking a

partial sum with 10 terms. (Compare with the exact value of the sum: 2 6 ). How many terms should we take

in the partial sum to have an error less than 0.0005?

Homework: What left from above, plus 1,3,4,8,12,13,15,18,19,24,25,26,29,30,33,34. E.C.: 35,36,37

15

9.4. Positive series: Other Tests:

The integral test is a very useful test, but it requires the verification of a few conditions, and sometimes it cannot

be applied.

Let us look at the series: 1

1

2 1nn

. This series can be analyzed with the integral test, but that will require some

effort.

However, this series looks similar with 1

1

2nn

, with which we’d like to compare it.

Theorem 1 (Ordinary Comparison Test): Consider 0 , for n na b n K . Then:

a) If 1

n

n

b

( C) then 1

n

n

a

( C).

b) If 1

n

n

a

( D) then 1

n

n

b

( D)

In words, a positive series smaller than a convergent series is convergent, greater than a divergent series is

divergent. Relate this to the observation about the danger of positive series to “diverge to infinity” when they

diverge.

The proof for a) is easy using bounded partial sums and monotonic sequence theorem. b) follows from a) using

contradiction.

For the example above: 1 1

2 1 2n n

, so

1

1

2 1nn

( C) .

Other examples: 1

1

2 1nn

,

1

1

1n n

,

1

1

1n n

,

21

1

1n n

,

21

1

1n n

,

21

5

2 4 3n n n

,

3 21

3 2

2 11n

n

n n

,

2

21

sin ( )

n

n

n

(some of these may be harder than they seem and may be solved much easier with the limit comparison test

below).

Using the ordinary comparison test and the p-series, we should be able to quickly prove the

convergence/divergence of all rational type series and of many other series. However, the ordinary comparison

test stops short of series when only a sign change happens, in series like 1

1

2 1nn

. For this, we give and we’ll

use the stronger test:

16

Theorem 2 (Limit Comparison Test):

Consider 0 and b 0n na and let (1) lim n

nn

aL

b .

If 0 L , then the series 1

n

n

a

and 1

n

n

b

have the same behavior ! (very strong);

If L=0 (think of this like n na b ) then if 1

n

n

b

( C) then 1

n

n

a

( C) ;

If L (think of this like n na b ) then if 1

n

n

b

( D) then 1

n

n

a

( D).

The proof for a) uses the definition of the limit with some ϵ and Theorem 1 after. Similarly for b) and c).

Remark: This is an easier to use test (be careful with the conditions) and it can be applied for almost all

examples and type of problems which require a comparison test. Exception for the given problems is 2

21

sin ( )

n

n

n

.

Other examples: 1

ln( )

n

n

n

, 2

1

ln( )

n

n

n

, 2

1

1

19n n n

,

2

1 !n

n

n

(this last can be compared with 2

1

1

n n

for

example).

(keep in mind that ln(n) is smaller (slower) than any power > 0 of n).

An even easier to use test for positive series is:

Theorem 3 (Ratio Test): Let 0na and assume that (2) 1lim n

nn

a

a

. Then:

If 1 , then 1

n

n

a

( C).

If 1 , then 1

n

n

a

( D).

If 1 , then the test is inconclusive.

Intuitively, this theorem makes sense when we think of (2) to say that na is a geometric series with ratio ρ,

for which the above conclusions are true. The proof follows these lines.

The ratio test is applied when there are many simplifications in 1n

n

a

a

and you expect 1lim 1n

nn

a

a

.

17

Examples: 2

1

1

n n

, 1

1

( 1)n n n

,

1

2

!

n

n n

, 2

1 !n

n

n

, 1

!n

n

n

n

, 3

1 (2 )!n

n

n

, 3

1

4 cos( )

n

n

n

-- use ordinary comparison test

Homework:

What left from above, plus 1,2,6,7,10,12,13,16,19,22,27,32. E.C.: 43 a,c, 44, 46

9.5. Alternating series. Absolute convergence and conditional convergence:

Most of the previous tests apply to positive series only. In order to extend our study to series with arbitrary

terms (positive and negative), we begin by studying in this section a special type of non-positive series, namely

alternating series.

Alternating series are series whose terms alternate signs.

Examples: 1

1

1 1 1 11 1

2 3 4

n

n n

; 1

1 2 3 41

2 3 4 5 1

n

n

n

n

A general alternating series is a series of the form : (1) 1

1

1n

n

n

b

, where 0, nb n K .

In (1) we usually assume that the first term is positive, if not, we can factor a – sign from the entire series.

Theorem 1 (Alternating Series Test):

Consider the alternating series 1

1

1n

n

n

b

, with 0, nb n K . If :

a) 1 1 is decreasing, that is , n n nb b b n K and

b) lim 0nn

b

Then 1

1

1n

n

n

b

is convergent. In this case, an estimate for the error nS S is 1n nS S b .

NOTE: The most important (and almost only condition) to check for an alternating series is that nb to be

decreasing. Condition b) is merely a necessary condition (remember divergence test, which has to be verified

also. We symbolize in short form the two conditions as 0nb . Note that the Alternating series test is a very

easy test to do, but it is a sufficient test only, and not necessary (it is inconclusive when the conditions above are

NOT MET)

18

Note the special form of an alternating series, which should be verified before you use the test.

Proof:

Figure 9.5.1. Terms of nS for an alternating series with 0nb

The terms of the partial sum nS for an alternating series under the conditions of the theorem are sketched In

Figure 9.5.1.

Note that 3 1 2 3 1 1( )S b b b S b since 2 3 0b b (since nb is decreasing). Similarly, 4 2S S .

In general, it is easily shown that 2 2 2n nS S ( 2nS is increasing) and

2 1 2 3 4 5 2 2 2 1 1( ) ( ) ( )n n nS b b b b b b b b . Therefore 2nS is convergent.

2 1 2 2 1 2 1, since 0n n n nS S b S b .

From the picture above, it is clear that 1n nS S S (or the other way around), so

1 1n n n n nS S S S S S b .

Q.E.D. ∎

Examples: Study the convergence of

1

1

1n

n n

. How many terms do we need to take in nS such that 0.001nS S ?

Study the convergence of

1

1

1

!

n

n n

. How close is 5S to S ?

The test above is easy to use, but quite limited. It does not say anything about general non-positive series (not

necessarily alternating), nor about the case when the conditions are not met.

Example: How do we study the series 2

1

sin( )

n

n

n

?

In order to deal with general series like the series above, we introduce a few notions.

19

Definition: A general (possibly non-positive) series 1

n

n

a

is said to converge absolutely (AC), if 1

n

n

a

converges.

(remember this as convergent in absolute value). Note that the second series is much easier to study, since it is a

positive series.

Theorem 2 (absolute convergence): If 1

n

n

a

is absolutely convergent (AC), then 1

n

n

a

is convergent ( C).

Proof: Consider n n n n n nv a a a v a . Since 0 2n nv a , and since 1

n

n

a

is ( C), then 1

n

n

v

( C).

Therefore

1

n

n

a

( C) (difference of two convergent series).

Theorem 2 is very useful when 1

n

n

a

is absolutely convergent (2

1

sin( )

n

n

n

), but it does not cover the cases when

it is not

(1

sin( )

n

n

n

). Of course, there are series which are not absolutely convergent, but just convergent.

Definition: If 1

n

n

a

is convergent but 1

n

n

a

is divergent, then we say that 1

n

n

a

conditionally convergent

(CC).

Examples: Study if each of the following series is (AC) or (CC). Hint: It is usually good to start with the

convergence of 1

n

n

a

(AC), to avoid unnecessary work).

1

1

1n

n n

,

1

1

1

!

n

n n

,

1

21

1n

n n

,

2

1 2nn

n

, 1

sin(4 )

4nn

n

, 1

cos( )

n

n

n

, 1

cos3

!n

n

n

The cases in which the series in not absolutely convergent (1

n

n

a

is divergent), but the series is not alternating,

are the hardest cases to study if the series is (CC) or (D). The following theorem can sometimes come to rescue

(it is much stronger than it seems). And it will be useful in the next sections also.

Theorem 3 (Absolute Ratio Test):

20

Let 1

n

n

a

be an arbitrary series (not necessarily positive). Consider 1lim n

nn

a

a

. Then:

If 1 then the series converges absolutely (this is just the ratio test for 1

n

n

a

…) so converges.

If 1 , 1

n

n

a

diverges (very useful and not obvious).

If 1 then the test is inconclusive.

The proof relies on lim 0 lim 0n nn n

a a

.

We will use Absolute Ratio Test sometimes for convergence, but be aware that when studying the convergence

of 1

n

n

a

, you are not limited at this test!

Examples: Classify each of the following series as (AC), (CC) or (D):

21

sin( )

n

n

n

, 1

1sin

n

nn

, 1

1 sinn

n n

(use AST for C),

1

41

1n

n n

,

1

1

1

ln( )

n

n n

(use AST for C)

2

1

1

1n

nn

n

e

, 1

1

11

( 1)

n

n n n

, 1

1

11

ln( )

n

n n n

Homework:

From above plus 1,2,3,6,7,13,14,15,18,23,26,30,32,40

This is a good place for a test.

21

9.6. Power Series:

In the previous sections of this Chapter, we have covered the important theory of series of numbers, that is

series of the form 1

n

n

a

, where na is a positive or arbitrary non-positive sequence. For the remaining sections,

we consider a more general type of series: power series, which are series of the form:

(1) 2 3

0 1 2 3

0

n n

n n

n

a x a a x a x a x a x

.

In (1), we convene that 0 1x , even for x=0. Note that the power series (1) reduces to a usual series of numbers

for each fixed value of x. Also, note that in most cases the power series (1) is non-positive, since x in general

can have negative values (and na itself can be non-positive).

The first question that we pose for a power series is:

For what values of x is the power series (1) convergent ?

When convergent, the sum of a power series of the form (1) is a function ( )f x , since for every 0x where the

series is convergent, the series converges to a distinct value 0( )f x . We write this in the form:

(2) 2 3

0 1 2 3( ) n

nf x a a x a x a x a x

In order to find the values of x where a power series converges, we always use the absolute ratio test.

Example 1: For what values of x are the following power series convergent?

0

! n

n

n x

Calculate 1 1

1 1 ( 1)!lim lim lim lim( 1) | | | |

!

n n

n n

n nn n n nn n

b a x n xn x x

b a x n x

.

Therefore, when 0x , 1 , and 0

! n

n

n x

(D) (see the absolute ratio test).

When 0x (that is 0x ), the series is considered 0

0

0! 1n

n

n

a x a

, and the series converges.

Therefore 0

! n

n

n x

for 0x and diverges otherwise.

22

The series 0

n

n

n

a x

is called a power series centered at 0. Most generally, we consider a power series centered

at some value a of the form: (3) 0

n

n

n

a x a

.

Example 2: For what values of x are the following power series convergent?

0

3n

n

x

n

Applying the absolute ratio test, we find that 3x . Therefore,

0

3n

n

x

n

is ( C) for

3 1 2 4x x

and (D) for 3 1 ( ,2) (4, )x x . In the endpoints x=2 and x=4 the ratio test is inconclusive,

and we study these cases separately, using the usual theory of series of numbers:

x=2:

0 0

3 1( )

n n

n n

xC

n n

(use the alternating series test)

x=4:

0 0

3 1( )

n

n n

xD

n n

(the harmonic series).

Therefore, the series

0

3n

n

x

n

converges for [2,4)x and diverges otherwise.

Theorem 1: For a general power series 0

n

n

n

a x a

there are three distinct possibilities:

The series 0

n

n

n

a x a

converges only for x a and diverges otherwise (see example 1);

The series 0

n

n

n

a x a

converges for all x R ;

There is a number R (called radius of convergence), such that the series converges for x a R

(possibly including one or both endpoints), and diverges for x a R (see example 3).

23

Proof:

For a general power series (3), apply the absolute ratio test to determine the interval of convergence:

1

11 1lim lim lim

n

nn n

n n nn n nn

a x ab ax a

b a x a a

. Denote 1 1

lim n

nn

a

a R

↔ 1

1

1/ lim limn n

n nn n

a aR

a a

(see

why below).

The conclusion follows from analyzing the three possibilities:

1lim 0n

nn

aR

a

(the series converges everywhere);

1lim 0n

nn

aR

a

(the series converges only for x a );

1lim 1/n

nn

av R v

a

(the series converges on ,a R a R possibly including the endpoints).

NOTE: You can use directly the formula 1

lim n

nn

aR

a

, although it is preferably to apply completely the

absolutely ratio test every time for clarity.

NOTE: The interval ,a R a R (possibly including one or both endpoints) is called the interval of

convergence of the power series. The first part of studying a power series is to determine its interval (radius) of

convergence. The function to which a power series convergence is studied after (and we will tackle that in the

following sections).

Examples: Solve problems 18, 23,5,6, in the text

Homework:

From above, plus problems 4,8,9,12,16,22,23,25,27,32,34,35. E.C. 36

24

9.7. Operations on Power Series:

Since we know how to find the interval of convergence for a general power series 0

n

n

n

a x a

, now we want

to answer the other important question for a power series:

(1) What is the function ( )f x which is the sum of a power series over its interval of convergence? (note

that we mentioned this at the beginning of the chapter when we said that “analytical functions” in your

calculator are expressed first as power series before being evaluated!).

In order to answer this question, we usually start with the opposite simpler question:

(2) Having an “analytical” (that is with infinitely many derivatives over its domain) function, can we write

this function as a power series? Answering this question will help us (hopefully) find the sum of a power

series by recognizing it as an expansion of a certain function (more on this later).

In this and the next section we give methods to answer question (2).

For many power series it is useful to start with the geometric series:

(3) 0

1

1

n

n

xx

, for 1x (the restriction 1x comes from the usual study of a geometric series such

that

1lim lim

1

n

nn n

xS R

x

). Convince yourself that this agrees with the interval of convergence of the power series

0

n

n

x

(absolute ratio test)!

Formula (3) can be regarded as the fundamental power series.

In this section, we show how to derive many new power series based on the formula (3). New power series can

be derived from (3) based on transformations of this (or combinations of them):

1. Substitution:

Since (3) is valid for ANY with 1x R x , then the variable x in this formula can be replaced with any

new expression (or variable) as needed. For example, we derive:

(4) 0

1

1

n

n

xx

for 1x

(5) 2

20

1

1

n

n

xx

for 2 21 1 1x x x

25

(6) 0

1 1n

n

bx

a bx a a

for 1

bx

a for any ,a b R and so on.

2. What about finding a power series for

2

1( )

1f x

x

? We could use

2 2( ) 1 1f x x x x x and deduce a general formula, but is there a simpler way?

Yes, we can take the derivative of the formula (3)

NOTE: If ( )f x has a power series expansion 0

n

n

n

a x a

on some convergence interval

,a R a R (possibly with endpoints), then '( )f x has the power series expansion 1

1

n

n

n

a n x a

on

,a R a R (possibly with endpoints). A similar results holds for integration of (3). This general result

is not very hard to prove: just answer the question: If the radius of convergence of 0

n

n

n

a x a

is R ,

what is the radius of convergence of 1

0 1

n n

n n

n n

da x a a n x a

dx

? Similarly for integration).

Taking the derivative of (3), we obtain:

(7)

2 3 2 1

21

1 11 ' 1 2 3

1 1

n

n

dx x x x x nx

dx x x

for 1x

(the derivative of the series but with n moved over by 1 since the derivative of the first term is 0).

(It is much clearer to take the derivative in the expanded form, at least for the beginning). Note that we

know ahead of time that the interval of convergence for the series in (7) is the same with the series in (3).

Why?

3. Integration: Can we find a power series expansion for ( ) ln(1 )f x x . Note that the domain of ( )f x is

,1 .

We remark that

(8) 2 3

2 3

0 1

1( ) ln 1 1

1 2 3

nn

n n

x x xf x x dx x dx x x x dx x

x n

for 1x .

Make the substitution 1y x in (8) to get:

26

(9) 1

1 1 1

1 ( 1) 1 ( 1) 1ln( )

n n nn n

n n n

y y yy

n n n

for 1 1 ?y y

Therefore we have derived a power series for the function ln. In a similar manner, we can derive power

series for other important function.

4. xe :

Consider the power series 2 3

( ) 11! 2! 3!

x x xS x You can easily find that the interval of convergence

for this power series is x .

Also, note that '( ) ( ), for all S x S x x . Solving this differential equation, with (0) 1S (remember

Section 6.5), we obtain ( ) xS x e . Therefore, we have found that:

(10) 2 3

( ) 11! 2! 3!

x x xS x for all x .

You should memorize some essential expansions (for xe , 1

1 x, but the method is more important).

5. Other operations:

We can derive power series by addition, subtraction, multiplication, (long) division, of known power

series.

Examples: Find the power series expansion of :

a) 2

4( )

1

xf x

x

b)

1( )

1

xf x ex

c) arctan( )

( )x

xf x

e d) 2 3( ) 1f x x x x

e) 2 3( ) 1f x x x x

Homework:

What left from above, plus 2,3,4,5,16,18,25,26. E.C.: 33,36-harder

27

9.8. Taylor and MacLaurin Series:

We consider now the more general problem: Given an analytical (with infinitely many derivatives over its

domain) function, can we find a power series expansion for this function around a point a in its domain.

First, let us assume that a power series exists for a given function. What is this power series?

Theorem 1 (uniqueness theorem):

Let :f D R be an analytical function over its domain D and let a D an arbitrary point.

Assume that ( )f x is expanded as a power series around a:

(1) 2 3

0 1 2 3( )f x c c x a c x a c x a on D

Then :

(2) ( ) ( )

!

n

n

f ac

n , where ( ) ( )nf a denotes the n’th derivative of the function ( )f x evaluated for x a .

(convention (0) ( ) ( )f a f a ).

NOTE: The above theorem says that is a function f(x) can be represented as a power series, this power series is

unique (the coefficients are uniquely determined by f).

Proof:

Let us prove (2) (let us find the coefficients of the power series), is we assume (1).

Evaluate (1) for x a to get: 0( )f a c . Take the derivative of (1) to get:

(3) 2

1 2 3'( ) 2 3f x c c x a c x a , and evaluate for x a to get 1'( )f a c , which gives 1c

(agrees with (2)).

Take the derivative of (3) to get:

(4) 2

2 3 4''( ) 2 2 3 4 3f x c c x a c x a and evaluate for x a to get 2 2

''( )''( ) 2

2!

f af a c c

Continuing this procedure it is clear that we obtain the formula (2) for the coefficients nc in the power expansion

of ( )f x around x a (this power series is called the Taylor expansion – or the Taylor series – of ( )f x around

x a .

A more rigorous proof can use mathematical induction to show (2).

Q.E.D.

28

Therefore, the general Taylor series of a function ( )f x around x a is:

(5) (3) ( )

2 3

0

'( ) ''( ) ( ) ( )( ) ( )

1! 2! 3! !

nn

n

f a f a f a f af x f a x a x a x a x a

n

Formula (5) is a very important formula which needs to be memorized (we’ll use it often).

Definition: The Taylor series of a function ( )f x around 0x is called the MacLaurin series (expansion) of

( )f x .

It has the form:

(6) (3) ( )

2 3

0

'(0) ''(0) (0) (0)( ) (0)

1! 2! 3! !

nn

n

f f f ff x f x x x x

n

.

Examples: Find the MacLaurin series for ( ) , ( ) ln( ), ( ) sin( ), ( ) cos( )xf x e f x x f x x f x x .

Do these agree with the formulas you have found before?

There is still a question about if the Taylor series exists for a given function (although we know what this series

is, when it exists). We have hinted that all analytical functions (functions which have infinitely many

derivatives) admit a Taylor series. The important Taylor’s theorem next states this:

Theorem 1 (Taylor’s theorem):

Let :f D R be a function with infinitely many derivatives on D and let a D be an arbitrary point (this

assumption can be weakened to say that :f D R has (n+1) derivatives around a). Then, for each x D (or

for each x in that neighborhood of a where f has (n+1) derivatives) we have:

(7) ( )

2'( ) ''( ) ( )( ) ( ) ( )

1! 2! !

nn

n

f a f a f af x f a x a x a x a R x

n , where the remainder (the

error)

( )nR x is

(8)

( 1)

1( )( )

1 !

nn

n

f cR x x a

n

, where c is some point between x and a .

NOTE: This theorem is important for two reasons:

It states that the Taylor series exists when ( )f x is analytical on D (or even when it has at least (n+1)

derivatives

around x=a);

More importantly, it gives an estimate for the error made when approximating the function only with a

finite number of terms in the expansion. In practice, we cannot calculate an infinity number of terms in

29

the Taylor expansion, and we want to know what is the size of the error made. However, note that the

new constant c is not known, all we know is that it is a constant between x and a (remember the mean

value theorem). We will have to deal with this.

Sometimes it is convenient to find the radius of convergence of the Taylor series in (7) R , and then we

can say that a R c a R which can be useful.

The proof of Taylor’s theorem is based on the Rolle’s theorem (try for n=2 following book) – do if time -- .

Examples:

Find the MacLaurin series above and agree with previous formulas. Memorize. Find MacLaurin series for

1

( ) , ( ) ln 1 , ( ) 1 , 1

pf x f x x f x x p R

x

.

Emphasize that they need to memorize the formula above, understand and memorize (5), (7) and (8) and

memorize them.

Homework:

What left from above plus 1,4,7,11,19,21,23,30,40

9.9. The Taylor Approximation Theorem:

Taylor’s theorem is a very important theorem in Calculus. Note that you met it in various particular forms

before.

For example, the definition of the derivative ( ) ( )

'( ) limx a

f x f af a

x a

is a particular form (replace ( )f x with its

Taylor expansion around a and see if it is true).

In this section we focus on analyzing the error in a Taylor (or MacLaurin) approximation for an

analytical function ( )f x .

What Taylor’s theorem says is that any analytical function is an infinite polynomial at any point in its domain.

Practically, it can be approximated by a (finite) polynomial at any points in its domain.

Looking at the formula for the error

( 1)

1( )( )

1 !

nn

n

f cR x x a

n

made in this approximation, we infer:

The error ( )nR x decreases by increasing n (taking more terms in the approximation) note that

approximating a function by a line at a point is rough (this is the approximation by a line with the slope

'( )f a at point:

( ) ( ) '( )f x f a f a x a ) ;

30

The error ( )nR x is relatively small when x is near a, but it gets large as x goes far from a (look at Figure

4 in book, try example

3 5 7

sin( )3! 5! 7!

x x xx x taking more and more terms, look how the error is small near x=0 but

larger and larger away from 0, in other words it is O.K. to consider this approximation when calculating

sin(0.5) say, but not O.K. when calculating 5

sin2

!

The error becomes bigger as ( )f x is “less smooth” (larger values of the derivative) over its domain.

Example:

Approximate 0.7e with an error less than 0.001 (do not just plug 0.7e in your calculator to get the result).

Solution:

Since 0.7 is close to 0, we can use the MacLaurin approximation: 2 3

12! 3! !

nx x x x

e xn

.

The main question is : how large should n be such that ( ) 0.001nR x ?

Since ( 1)

1 1( )( )

( 1)! ( 1)!

n cn n

n

f c eR x x x

n n

→

1(0.7) 0.7

( 1)! ( 1)!

c cn

n

e eR

n n

.

However, remember that 0.70 0.7 1 3cc e e →

3(0.7) 0.001

1 !nR

n

.

By trial and error, we see that 6n , for which we find 6 (0.7)T which gives an approximation of 0.7e with an

error less than 0.001.

Homework:

5,6,15,19-28 (choose 5), 31,38,39,42,43,56,62

CHAPTER REVIEW: ALL