chapter infinite sequences and...

CHAPTER 10 INFINITE SEQUENCES AND SERIES

10.1 Sequences 10.2 Infinite Series 10.3 The Integral Tests 10.4 Comparison Tests 10.5 The Ratio and Root Tests 10.6 Alternating Series: Absolute and Conditional Convergence 10.7 Power Series 10.8 Taylor and MacLaurin Series

Calculus & Analytic Geometry II (MATF 144) 2

10.1 Sequences

Definition

An infinite sequence or more simply a sequence is an unending succession of

numbers, called terms.

It is understood that the terms have a definite order, that is, there is a first term

a1, a second term a2, a third term a3, and so forth.

Such a sequence would typically be written as

a1, a2, a3, a4,….an…

where the dots are used to indicate that the sequence continues indefinitely.

Some specific example are

1, 2, 3, 4, ….. 1 1 1

1, , , ,....2 3 4

2, 4, 6, 8, ….. 1, ‐1, 1, ‐1,….

The number an is called the nth term or general term, of the sequence.


Example 10.1:

In each part, find the general term of the sequence.

(a) 1 2 3 4, , , ....

2 3 4 5 (b) 1 1 1 1, , , ....

2 4 8 16

(c) 1 2 3 4, , , ....

2 3 4 5

(d) 1, 3, 5, 7, …..


When the general term of sequence is known, there is no need to write out the initial

terms, and it is common to write the general term enclosed in braces.

Sequence Brace Notation

1 2 3 4, , , .... ....

2 3 4 5 1

n

n 11

n

n

n

n

1 1 1 1 1, , , .... ...

2 4 8 16 2n 1

1

2

n

nn

11 2 3 4, , , ....,( 1) ,....

2 3 4 5 1n n

n

1

1

( 1)1

nn

n

n

n

1, 3, 5, 7, …..,2n‐1,… 12 1

nn

n


Convergence and Divergence

Sometimes the numbers in a sequence approach a single value as the index n increases.

This happens in the sequence

1,12,13,14, … ,

1, …

whose terms approach 0 as n gets larger, and in the sequence

0,12,23,34,45, … ,1

1, …

whose terms approach 1.

On the other hand, sequences like

√1, √2, √3,… . . , √ , … .

have terms that get larger than any number as n increases, and sequences like

1, 1,1, 1… . , 1 , … .

bounce back and forth between 1 and ‐1, never converging to a single value.


Example 10.2:

Find the limit of each of these sequences.

(a)

1

n

(b) ( 1)n (c) 8 2n


Calculating Limits of Sequences


Example 10.3:

Determine whether the sequence converges or diverges. If it converges, find the limit.

(a)

100

n

(b)

1n

n

(c)

4

4 2

3 1

5 2 1

n n

n n

(e)

2

3

2 5 7n n

n

(b)

5 3

4 2

2

7 3

n n

n n

(b) 1( 1)2 1

n n

n


The Sandwich Theorem

Example 10.4:

Find the limit of the sequence:

(a) 2

sinn

n

(b)

2cos

3nn

(c)

1( 1)n

n


Using L’Hôpital’s rule.

The next theorem enables us to use L’Hôpital’s rule to find the limits of some sequences. It

formalizes the connection between lim nn

a

and lim ( )x

f x .

Example 10.5:(Evaluating a limit using L’Hôpital’s rule).

(a)

lnlimn

n

n (b)

2

lim2nn

n

(c) n

nn /1lim

(d)

1lim

1

n

n

n

n


Commonly Occuring Limits

The next theorem gives some limits that arise frequently.


Example 10.6:

Find the limit of each convergent sequence.

(a)

lnn

na

n

(b) 3n

na n (c)

25nna n

(d)

1

3

n

na (e)

3n

n

na

n

(f)

55

!

n

nan


10.2 Infinite Series


i) The nth partial sum Sn of the series na is

1 2 ...n kS a a a ii) The sequence of partial sums of the series na is

1 2 3, , ,..... ,...nS S S S

Example 10.7:

Given the series

1 1 1 1... ...

1 2 2 3 3 4 ( 1)n n

(a) Find S1, S2, S3, S4, S5 and S6.

(b) Find Sn.

(c) Show that the series converges and find its sum.


Example 10.8: Given the series,

1 1

1

( 1) 1 ( 1) 1 ( 1) ... ( 1) ...n n

n

(a) Find S1, S2, S3, S4, S5 and S6.

(b) Find Sn.

(c) Show that the series diverges.


Geometric Series

Example 10.9:

Determine whether each of the following geometric series converges or diverges. If

the series converges, find its sum.

(a) 0

1 3

7 2

n

n

(b) 0

13

5

n

n


Telescoping Series

A telescoping series does not have a set form, like the geometric series do. A telescoping

series is any series where nearly every term cancels with a preceding or following term. For

instance, the series

21

1

k k k

Using partial fractions, we find that

2

1 1 1 1

( 1) 1k k k kk k

Thus, the nth partial sum of the given series can be represented as follows:


21 1

1 1 1

1

1 1 1 1 1 1 11 ..

2 2 3 3 4 1

1 1 1 1 1 1 11 ..

2 2 3 4 1

11

1

n n

nk k

Sk kk k

n n

n n n

n

The limit of the sequence of partial sums is

1 1lim lim 1

1n

n nS

n n

so the series converges, with sum S = 1.


Example 10.10:

In each case, express the nth partial sums Sn in terms of n and determine whether

the series converges or diverges.

(a) 0

1

( 1)( 2)n n n

(b) 0

1 1

2 1 2 3n n n


Divergent Series

One reason that a series may fail to converge is that its terms don’t become small. For

example, the series

2 2

1

1 4 9 .... ...n

n n

This series is diverges because the partial sums grow beyond every number L. After n = 1,

the partial sums 21 4 9 ....ns n is greater than

2n .

Theorem above states that if a series converges, then the limits of its nth term an as n

is 0.Sometimes, it is possible for a series to become diverges.


lim 0nn

a

So this theorem leads to a test for detecting the kind of divergence that occurred in some of the series.

The nth‐Term Test

If , then further investigation is necessary to determine whether the series 1

nn

a

is converge or diverge.


Example 10.11:

Applying the nth‐Term Test

(a) 12 1n

n

n

(b) 2

1n

n

(c) 1

1

( 1)n

n


Combining Series


Example 10.12:

(a) 11

7 2

( 1) 3nn n n

(b) 1 11

1 1

2 6n nn

(c)

2

1

1 1

2 3n nn


10.3 The Integral Tests


Example 10.13:

Determine whether the following series converge.

(a) 21 1n

n

n

(b) 1

1

2 5n n

(c)

0

ln

n

n

n

The p‐Series The Integral Test is used to analyze the convergence of an entire family of infinite series

known as the p – series. For what values of the real numbers p does the p – series

converge? 1

1p

n n


Example 10.14:

Test each of the following series for convergence.

(a) 101

1

n n

(b) 31

1

n n

(c) 24

1

( 1)n n


10.4 Comparison Tests

We have seen how to determine the convergence of geometric series, p‐series and a few

others. We can test the convergence of many more series by comparing their terms to

those of a series whose convergence is known.

Note:Let na , nc and nd be series with positive terms. The series na converges if it is “smaller” than(dominated by) a

known convergent series nc and diverges if it is “larger” than (dominates)a known divergent series nd . That is, “smaller than

convergent is convergent”, and “bigger than divergent is divergent”.


Example 10.15:

Test the following series for convergence.

(a) 1

1

3 1nn

(b) 2

1

1n n

(c) 32

ln

n

n

n


Example 10.16:

Test the following series for convergence by using the limit comparison test.

(a) 1

1

2 5nn

(b) 1

3 2

(3 5)n

n

n n

(c)

1

100

70nn

n

ne


10.5 The Ratio and Root Tests

Intuitively, a series of positive terms na converges if and only if the sequence na

decrease rapidly toward 0. One way to measure the rate at which the sequence na is

decreasing is to examine the ratio nn aa /1 as n grows large. This approach leads to the

following theorem:


Note: You will find the ratio test most useful with series involving factorials or

exponentials.

Example 10.17:

Use the ratio test to determine whether the following series converge or diverge.

(a)

1 !

2

n

n

n (b)

1 !n

n

n

n

(c) 1

n

n

ne

(d)

1

( 1)( 2)

!n

n n

n


The following test is often useful if an contains powers of n.


Example 10.18:

Determine whether the series is convergent or divergent.

(a)

1 )(ln

1

nnn

(b) 21

2n

k n

(c) 1 2 1

n

n

n

n


Guidelines to Use the Convergence Test.

Here is a reasonable course of action when testing a series of a positive terms

1nnafor

convergence.

1. Begin with the Divergence Test. If you show that lim 0nn

a

, then the series diverges

and your work is finished.

2. Is the series a special series?Recall the convergence properties for the following

series.

Geometric Series: converges if 1 and diverges for 1.nar r r

p‐series: 1 converges for 1 and diverges for 1.

pp p

n

Check also for a telescoping series.


3. If the general nth term of the series looks like a function you can integrate, then try

the Integral Test.

4. If the general nth term of the series involves, where a is a constant, the Ratio Test is

advisable. Series with n in an exponent may yield to the Root Test.

5. If the general nth term of the series is a rational function of n (or a root or a rational

function), use the Direct Comparison or the Limit Comparison Test. Use the families of

series given in Step 2 as a comparison series.


10.6 Alternating Series, Absolute and Conditional

Convergence

Alternating Series

We consider alternating series in which signs strictly alternate, as in the series,

1

1

1 1 1 1 1 1 1 11 ...

2 3 4 5 6 7 8

n

n n

The factor 11

n has the pattern {…,1,‐1 ,1, ‐1, …} and provides the alternating signs.

We prove the convergence of the alternating series by applying the Alternating Series Test.


Example 10.19:

(a)

1

21

( 1)n

n n

(b)

12

1

2( 1)

4 3n

n

n

n

(c)

1

1

2( 1)

4 3n

n

n

n


Absolute and Conditional Convergence

The convergence test we have developed cannot be applied to a series that has mixed

terms or does not strictly alternate. In such cases, it is often useful to apply the following

theorem.

The series

1

21

( 1)n

n n

is an example of an absolutely convergent series because the series

of absolute values,

1

2 21 1

( 1) 1n

n nn n

is a convergent p‐series. In this case, removing the alternating signs in the series does not

affect its convergence.


On the other hand, the convergent alternating harmonic series

1

1

( 1)n

n n

has the

property that the corresponding series of absolute values,

1

1 1

( 1) 1n

n nn n

does not converge. In this case, removing the alternating signs in the series does effect

convergence.


Example 10.20: Determine whether the following series diverge, converge absolutely, or

converge conditionally.

a)

1

1

( 1)n

n n

b)

1

31

( 1)n

n n

c) 1

sin

n

n

n

d) 1

1( 1)

2n

n

n

n




10.7 Power Series

The most important reason for developing the theory in the previous sections is to

represent functions as power series‐that is, as series whose terms contain powers of a

variable x.

The good way to become familiar with power series is to return to geometric series.

Recall that for a fixed number r,

2

0

11 ... , provided 1

1n

nr r r r

r

It’s a small change to replace the real number r by the variable x. In doing so, the geometric

series becomes a new representation of a familiar function:

2

0

11 ... , provided 1

1n

nx x x x

x


Convergence of Power Series

We begin by establishing the terminology of power series.



Example :

Find the interval and radius of convergence for each power series:

a) 0 !

n

n

x

n

Solution: The center of the power series is 0 and the terms of the series are. We test the series for absolute convergence using Ratio Test:

1

1

/( 1)!lim (Ratio Test)

/ !

! lim (Invert and multiply)

( 1)!

1 lim 0 (Simplify and take the limit with fixed)

( 1)

n

nn

n

nn

n

x n

x n

x n

nx

x xn


Notice that in taking the limit as n , x is held fixed. Therefore, 0 for all values

of x, which implies that the interval of convergence of the power series is

x and the radius of convergence is R


b) 0

( 1) ( 2)

4

n n

nn

x

Solution: We test for absolute convergence using the Root Test:

(‐1) ( 2)lim (Root Test)

4

2 1

4

n n

nnn

x

x

In this case, depends on the value of x. For absolute convergence, x must satisfy

21

4

x

which implies that 2 4x . Using standard techniques for solving inequalities, the

solution set is 4 2 4, or 2 6x x . Thus, the interval of convergence

includes (‐2, 6).


The Root Test does not give information about convergence at the endpoints, x = ‐2

and x = 6, because at these points, the Root Test results in 1 . To test for

convergence at the endpoints, we must substitute each endpoint into the series and

carry out separate tests.

At x = ‐2, the power series becomes

0 0

0

( 1) ( 2) 4 Substitute 2 and simplify

4 4

1 Diverges by nth term test/divergence test

n n n

n nn n

n

xx

The series clearly diverges at the left endpoint. At x = 6, the power series is

0 0

0

( 1) ( 2) 41 Substitute 6 and simplify

4 4

1 Diverges by nth term test/divergence test

n n nn

n nn n

n

n

xx


This series also diverges at the right endpoint. Therefore, the interval of convergence

is ( ‐2, 6), excluding the endpoints and the radius of convergence is R =4.

(When the convergence set is the entire x‐axis).

Show that the power series 1 !

n

n

x

n

converges for all x.


Example 10.21:

(Convergence only at the point x=0).

Show that the power series 1! n

nn x

converges only when x = 0.


Example 10.22:

(Convergence set is a bounded interval).

Find the convergence set for the power series 1

n

n

x

n

.


According to the Theorem 11.23, the set of numbers for which the power series 0

nn

na x

converges is an interval centered at x =0. We call this the interval of convergence of the

power series. If this interval has length 2R, then R is called the radius of convergence of the

series. If the series has radius of convergence R =0, and if it converges for all x, we say that

R .


Example 10.23:

Find the interval of convergence for the power series 0

2n n

n

x

n

.

What is the radius of the convergence?

Example 10.24:

Find the interval of convergence of the power series 0

( 1)

3

n

nn

x

.


Term‐by‐term Differentiation and Integration.

Suppose that a power series 0

nn

na x

has a radius of convergence r>0, and let f be defined

by

2 30 1 2 3

0( ) ... ...n n

n nn

f x a x a a x a x a x a x

for every x in the interval of convergence. If –r < x < r, then

(i) 2 1 1

1 2 31

( ) 2 3 ... ...n nn n

nf x a a x a x na x na x

(ii)

2 3 11

0 0 1 21

( ) ... ...2 3 1 1

nnx n

nn

x x x af t dt a x a a a x

n n


Example 10.25:

Find a function f that is represented by the power series

2 31 ... ( 1) ...n nx x x x

Example 10.26:

Find a power series representation for 2

1

(1 )x if 1x .


Example 10.27:

Find a power series representation for ln(1 )x if 1x .

Example 10.28:

Find a power series representation for1tan x

.


10.8 Taylor and MacLaurin Series

In the previous lecture, we considered power series representation for several special

functions, including those where f(x) has the form

1

(1 )x , ln(1 )x and 1tan x

,

provided x is suitably restricted. We now wish to consider the following question.

If a function f has a power series representation

0( ) n

nn

f x a x

or 0( ) ( )nn

nf x a x c

what is the form of na ?

Suppose that,

2 3 40 1 2 3 4

0( ) ...n n

nf x a x a a x a x a x a x


and the radius of convergence of the series is r > 0. A power series representation for ( )f x

may be obtained by differentiating each term of the series for f(x). We may then find a

series for ( )f x by differentiating the terms of the series for ( )f x . Series for ( )f x , (4)( )f x

and so on, can be found in similar fashion. Thus,

2 3 11 2 3 4

1( ) 2 3 4 ... n

nn

f x a a x a x a x na x

2 22 3 4

2( ) 2 (3 2) (4 3) ... ( 1) n

nn

f x a a x a x n n a x

33 4

3( ) (3 2) (4 3 2) ... ( 1)( 2) n

nn

f x a a x n n n a x

and for every positive integer k,

( )( ) ( 1)( 2)...( 1)k n kn

n kf x n n n n k a x


Each series obtained by differentiation has the same radius of convergence r as the series

for f(x). Substituting 0 for x in each of these representations, we obtain

0 1 2 3(0) , (0) , (0) 2 , (0) (3 2)f a f a f a f a

and for every positive integer k,

( )( ) ( 1)( 2)...(1)kkf x k k k a

If we let k =n, then

( )( ) !nnf x n a

Solving the preceding equations for ,....,,, 210 aaa we see that

0 1 2 3

(0) (0)(0), (0), ,

2 (3 2)

f fa f a f a a

And, in general,

( )(0)

!

n

n

fa

n


MacLaurin Series for f(x)


0( ) n

nn

f x a x

with radius of convergence r > 0, then ( )(0)kf exists for every positive integer k and

( )(0)

!

n

n

fa

n

. Thus,

( )2

( )

0

(0) (0)( ) (0) (0) ... ...

2! !

(0)

!

nn

nn

n

f ff x f f x x x

n

fx

n


Taylor Series for f(x)


with radius of convergence r > 0, then ( )( )kf c exists for every positive integer k and

( )( )

!

n

n

f ca

n

. Thus,

2

( )

( )

0

( )( ) ( ) ( )( ) ( ) ...

2!

(0)( ) ...

!

( )( )

!

nn

nn

n

f cf x f c f c x c x c

fx c

n

f cx c

n

0( ) ( )nn

nf x a x c


Example 10.29:

Find the MacLaurin Series for f(x) = cos x.

Example 10.30:

Find the Taylor Series for f(x) = ln x at c =1.

Prepared by: Pn.Suriawati Sahari

chapter infinite sequences and...

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