1

UNIT 4 INFINITE SERIES Unit Structure

4.0 Overview

4.1 Distinguish between sequences and series

4.2 Sequences

4.3 Series

4.3.1 Convergence and divergence

4.3.1.1 The partial sum criterion

4.3.1.2 Geometric Series

4.3.1.3 Divergent Series

4.3.1.4 The Nth - Term Test for Divergence

4.3.1.5 The p-Series test

4.3.1.6 Comparison Test for Convergence

4.3.1.7 Limit Comparison Test

4.3.1.8 The Ratio Test

4.4 Strategies for Series

4.5 Power Series

4.6 Taylor Series and MacLaurin Series 4.6.1 Taylor Theorem

4.6.2 Taylors Formula for Functions of Two Variables

4.7 Summary

4.8 Answers to Activities

2

4.0 OVERVIEW In this chapter we study sequences and infinite series. Series play an important role in the field of ordinary differential equations and without series large portions of the field of partial differential equations would not be possible. 4.1 DISTINGUISH BETWEEN SEQUENCES AND SERIES Sequences? Series? Avoid the mess !!!!!!!! Wrong ! Right! Sequences General sequence terms are denoted as follows, { }KK ,a , ,a ,a ,a k321 In the notation above we must be cautious with the subscripts. The subscript of n+1 denotes the next term in the sequence and NOT one plus the nth term! In other words,

1 a a n1 n ++ So, when writing subscripts to make sure that the +1 does not migrate out of the subscript!

No need to bother, Same Stuff!

A sequence is a list of numbers written in a specific order whereas a series is the summation of a list of numbers or sequence.

3

Series Consider a sequence

{ } { } ,a , ,a ,a ,a a k3211 nn KK==

and then add up all the terms of the sequence, so as to get a series :

=

=+++++1 n

nk321 a a a a a KK .

Thus, as mentioned previously, a series is the summation of a list of numbers or sequence. Since we started out with an infinite sequence we sum up an infinite list of numbers. Because of this the series above is sometimes called an infinite series. The n is often called an index of summation or just index for short. Examples of sequences are :

=

+

= ,

256 ,

165 ,

94 ,

43 2,

n1n

1 n2 K ,

( )

=

=

+ ,

161 ,

81 ,

41 ,

21 1,-

21

0 nn

1 n

K

and

{ } == ofdigit n b where, b ht n1 nn

Remark :

in the first two series, to get the first few terms, we need to plug in values of n into the corresponding formula. This sequence is different from the first two in the sense that it does not have a specific formula for each term. However, it does tell us what each term should be. Each nth term should be the nthdigit of . We know that = 3.14159265359 The sequence is then { }K 9, 5, 3, 5, 6, 2, 9, 5, 1, 4, 1, 3,

4

4.2 SEQUENCES Definitions 1. We say that

L a lim nn =

if we can make na as close to L as we want for all sufficiently large n. In other words, the value of the na s approach L as n approaches infinity.

2. We say that

= a lim nn

if we can make na as large as we want for all sufficiently large n. That is, the value of the na s get larger and larger without bound as n approaches infinity.

3. We say that

= - a lim nn

if we can make na as negative as we want for all sufficiently large n. That is, the value of the na s get negative and larger without bound as n approaches infinity.

4. If nn a lim exists and is finite we say that the sequence is convergent. If nn a lim

does not exist or is infinite we say the sequence diverges.

Next we investigate how to find the limits of sequences. We will use the following theorems and facts.

Theorem 1

Given the sequence { }na if we have a function ( )xf such that ( ) na nf = and ( ) L xf limx

= then L a lim nn = . This theorem basically tells us that we take the limits of sequences much like we take the limit of functions.

5

Theorem 2

If 0 a lim nn = then 0 alim nn =

This theorem is convenient for sequences that alternate in signs and note that it will only work if the sequence has a limit of zero.

Theorem 3

The sequence { }=1 nnr converges if 1r 1

6

Facts Let {an} and {bn} be convergent sequences and let c be a real number. Then ( ) nnnnnnn b lim a lim b alim +=+

nnnn

a limc a c lim = ( ) ( ) ( )nnnnnnn b lim a lim b alim =

,b lim

a lim

balim

nn

nn

n

n

n

= provided 0 b lim nn Example 1. Determine if the following sequences converge or diverge. If the sequence converges then determine its limit.

(a)

=

+

2 n22

2

n 5 n 101n 3

(b)

=

1 n

n 2

ne

(c) ( )

=

1 n

n

n1-

7

Solution (a) We factorize the largest power of n from the numerator and denominator and then

take the limit. Thus, we get

n5n 101n 3lim 2

2

n +

5

n10

n1 3

lim 5

n10n

n1 3n

lim n2

2

n +

=

+

=

22

53 =

(b) Using Theorem 1,

x

elim n

elim x2

x

n 2

n = ,

Applying LHopitals rule, we get

==

1e 2lim

xelim

x2

x

x2

x

So, the sequence diverges.

(c ) Using Theorem 2,

( ) ( )n1- lim

n1- lim

n

n

n

n =

and the right hand side of the latter equation can be written as n1 lim

n . This limit

is obviously zero. Therefore, ( ) 0 n1- lim

n

n= and hence the corresponding sequence

converges to zero.

8

Example 2 Find the following limits

(a) ( ) n

1- limn

n

(b) n

cos(n) limn

(c) ( ){ }=1 nn1- Solution (a) This sequence converges to zero by Theorem 2. (b) Recall that ( ) 1 n cos 1 ,

for all integers n. Hence,

( )n

1 n cos n1 .

Since 0 n

1 lim n1lim

nn== , applying Theorem 4, ( ) 0 n cos lim n =

(c) By Theorem 3, this sequence converges to 1 . Activity 1 Find the limit of the following converging sequence :

(a)

=

+1 n

n

n 1 ( 0 and 0 )

(b)

=

1 n

n

n2n

9

4.3 SERIES 4.3.1 CONVERGENCE AND DIVERGENCE 4.3.1.1 THE PARTIAL SUM CRITERION Consider the following sequence { } { } , ,S , ,S ,S ,S S k3211 jj KK== , where k321

1 n a a a a a ++++==

=K

k

nkS .

The sequence { }=1 jjS is said to be a sequence of partial sums of the infinite series : KK a a a a a k321

1 nn +++++=

=

If the sequence of partial sums is convergent and if we define, S Slim kk = , then

S a1n

n ==

If the sequence of partial sums is divergent (i.e. either the limit does not exist or is infinite) then we call the series divergent.

10

4.3.1.2 GEOMETRIC SERIES

A geometric series may be written in the form = 0 n

nr a and its partial sum can be written

( )r1r-1a S

n

n = . We now discuss the convergence of a geometric series based on the magnitude of the common ratio r. The series will converge provided the partial sums form a convergent sequence. We therefore take the limit of the partial sum, which can be written as follows : ( )

r1r a -

r1a

r1r-1a S

nn

n == , so that ,

r1r alim -

r1alim S lim

n

nnnn =

nn

r lim r1

a - r1

a = . Now, from Theorem 3 we know that the above limit will exist and be finite provided -1 < r < 1. ( r can not be 1 since this leads to division by zero.)Therefore,

r1a S lim nn = .

Hence, a geometric series will converge if -1 < r < 1 or 1 r < Example 3 Determine the convergence of the following series.

(a) =

+

0 n n

nn

97 2

(b) = 0

n 2-n e 4n

(c ) =

+

0 3 n 4

6 - 1 -n 4

6

n

11

Solution

(a) We express the series as the sum of two geometric series which are convergent. Hence, the series is convergent. (The sum of two convergent series is also a convergent one.)

97

9 2

97 2

0 n

n

0 n

n

0 n

nn =

=

=

+

=

+

n

97 - 1

1

92 - 1

1 +=

(b) = 0

n 2-n e 4n

= =

=

0 2

n

2

e4 - 1

1 e4

n

, which is a convergent geometric series.

(c ) 3 k 4

6 - 1 -k 4

6 S0

n =

+=

n

k

is series is a telescoping series with partial sums:

Thus, every term except the first and last term canceled out. This is the origin of the name telescoping series.

3n 46 - 2 Sn +=

We determine the convergence of this series by taking the limit of the partial sums:

2 3 n 4

6 - 2lim Slim n

n n

=

+=

The sequence of partial sums is convergent and so the series is convergent and has a value of 2. That is,

2 3 k 4

6 - 1 -k 4

6 0

=

+

=k

( ) ( )

++++

+++

+

+

=3 1 n 4

6 1 - 1 n 4

6 3 n 4

6 1 -n 4

6 166 -

116

116 -

76

76 -

36 Sn K

12

Activity 2 Determine the convergence of the following series.

(a) =

++

0 n 2 2 n 3 n

1

(b) =

++

0 n 2 3 n 4 n

1

(c) e 7 - 3 n 4 n

1

0 n

n 5-n2

=

++

13

4.3.1.3 DIVERGENT SERIES Geometric series with r 1 are not the only series to diverge.

The series =1n n

1 is known as the Harmonic Series and this series diverges to .

We show graphically that the series is divergent.

Consider the graph of yx

= 1 on the interval [ ) , 1 as shown in Fig. 1.

y

x

y =1

y = 1 2

y = 1 3 y = 1

N

1 2 3 N N+ 1

Figure 1 The partial sum

=

++++==N

1nN N

1....31

21

11

n1S

= sum of areas of shaded rectangles

area under curve yx

= 1 from 1 to N+1

= [ ] 1N11N1 xlndxx1 ++ =

= ( ) ( ).1Nln1ln1Nln +=+ We find that ( )1NlnSN +

14

Since ( ) ++1Nln as ,N + we find that ( )

= +++=+=

1nN

NN

1Nlnlimn1Slim

Since the sequence of partial sums { }NS diverges, the Harmonic Series is a divergent series. 4.3.1.4 THE NTH - TERM TEST FOR DIVERGENCE

If 0ulim nn

or if nn ulim fails to exist, then

=1nnu diverges.

This test only says that a series is guaranteed to diverge if the series terms do not go to zero in the limit. If the series terms do happen to go to zero the series may or may not converge!

The condition that if =1n

nu converges, then 0ulim n n

= is thus a Necessary

condition for convergence and not sufficient condition for the series=1n

nu to converge.

To see that 0ulim n

n=+ is not a sufficient condition for convergence, we consider the

Harmonic Series =1n n

1 .

We have 0n1lim

n=+ , but the series

=1n n1 diverges.

Example 4 Determine the convergence of the following series.

(a) = +1 n

n

2 12 - 1

n

(b) = +1n 1n

n , for 0

(c)

= n

1sin n1n

15

Solution

(a) Since 1, 1

21

1 - 21

lim 2 12 - 1lim

n

n

n n

n

n =

+=

+ the series diverges by the nth term test.

(b) By division, we get

n + 1 1

n n + 1

- 1

Therefore

+=+ 1n 1

1lim

1n nlim

nn

= 1 since 0

1n 1 + as n .

Using the nth - Term Test for divergence, = +1n 1n

n diverges since 01n

nlimn

+ .

(c)

n1

n1sin

lim n1sin n lim

n n

=

,

applying LHopitals rule, we obtain Hence, the series diverges.

1

n1-

n1 cos

n1-

lim

n1

n1sin

lim

2

2

n n =

=

16

Activity 3 (a) Determine the convergence of the following series.

= +1n 3

32

n 2 10n - n 4

(b) Consider the following series :

=

+

+1n

n2

x3 n 2

1 n

(i) Using the nth-term test for divergence, show that the series diverges when x = 1.

(ii) Can you use the same test to conclude about the nature of the series with 1 x < ?

Justify your answer.

(c) = +1 n 2 6 n 7

n 5

(d) =1 n

n 7

4.3.1.5 THE P-SERIES TEST

Theorem 5 If p is a real constant, the series =1 pn

1

n

converges if p>1 and

diverges if p 1.

17

4.3.1.6 COMPARISON TEST FOR CONVERGENCE

Let =1n

nu be a series such that 0>nu for all n . (a) Test for Convergence Suppose nn vu0 < for all 0nn > where 0n is some positive integer. If

=1nnv converges, then

=1nnu converges.

(b) Test for Divergence

Suppose nn uv0 < for all 0nn > . If =1n

nv diverges, then =1n

nu diverges.

In other words, we have two series of positive terms and the terms of one of the series is always larger than the terms of the other series. If the larger series is convergent then the smaller series must also be convergent. Likewise, if the smaller series is divergent then the larger series must also be divergent. Do not misuse this test. Just because the smaller of the two series converges does not say anything about the larger series. The larger series may still diverge. Likewise, just because we know that the larger of two series diverges we can not say that the smaller series will also diverge! Example 5 Determine if the following series converge or diverge.

(a) ( )

= 1 22 ncosnn

n

(b) = +

+1n

4

2

5 n3 n

18

Solution

(a) For n large, ( ) n1

ncosnn

22 and also, ( ) n

1 ncosn

n22

> .

Further, =1 n

1

n is a harmonic series and is therefore divergent. Hence, by the

comparison test, ( )

= 1 22 ncosnn

n diverges (or by theorem 5 , the p-series test).

(b) Observe that 4

2

4

2

n3 n

5 n3 n +

19

4.3.1.7 Limit Comparison Test

Suppose that 0u n > and 0vn > for all 0n > and that = nn

n vu

lim L .

(i) If

20

(a) We compare the series with =1n n

n

75 . Note that

n - 77 lim

75

n - 75

limn

n

n n

n

nn

n = .

The latter limit is clearly an indeterminate case and we therefore use LHopitals rule to simplify it.

1 - 7ln 77ln 7 lim

n - 77 lim

n

n

n n

n

n = . Since we get another indeterminate case, we apply LHopitals rule again to obtain

( )( ) 1 7ln 7

7ln 7 lim 1 - 7ln 7

7ln 7 lim 2n

2n

n n

n

n == .

So that,

1

75

n - 75

limn

n

nn

n = .

=1n n

n

75 is a convergent geometric series and therefore, by the limit comparison

test, = 1n n

n

n 75 is convergent.

(b) We compare the series with =2n n

1 . The ratio of the general terms for large n is

given by

nln n lim

n1

nln 1

lim n n = .

Using LHopitals rule, we get === n lim nln n lim

n1

nln 1

lim n n n

.

(c) For large n, nn

n u4

3n += behaves like n

1 vn = . =1n

nv diverges and

34

4

4

3

n

n

n1 1

1 n n

n n1

n nn

vu

+=+=+= > n as 0 1 . Hence,

= +1n 43

n nn diverges.

21

(d) For large n, nn 1

n n u2n +

+= behaves as 23n n

1 v = . =1n 23n

1 converges (p-series

test0 . >+

+= n as 0 1

n1 1

n1 1

vu

25

21

n

n . Hence, = +

+1n

2 nn 1n n converges.

Activity 5

Use the limit comparison test to determine the convergence of the series

(a) = +2n 2 1 n

1

(b) =

1n 1 -n 1 -

n 1

Hint : Compare it with =1n n

1 .

(c) =

1n

n1

1 - e

( Hint : compare it with =1n n

1 .

Indeed, for large n, the MacLaurin series of 2

n1

n21

n1 1 - e + . We neglect the

term2n 2

1 )

(d) =1n

sin 2n

1

22

4.3.1.8 THE RATIO TEST

Let =1n

nu be a series with 0u n > and suppose that =+ n1n

n uu

lim L .

Then the series (a) converges if L < 1 (b) diverges if L > 1. (c) may converge or it may diverge if L = 1. Example 7

(a) Consider the series 321

n

n n=

. Solution Here

2

nn

n3u = and ( )2

1n1n

1n3u+

=+

+

( ) ( ) ( )22

2n2

1n

2

n

2

1n

n

1n

1n3nn

.31n3

n3

1n3

uu

+=

+=

+=

+++

= 12nn

n32

2

++

=

2n1

n21

3

++

Now,

++

= +

2

nn

1nn

n1

n21

3limU

Ulim = 3. Since L = 3 > 1, the series

=1n 2n

n3 diverges.

23

(b) Next, consider the series ( )( )

=1n

2

! 2n !n

Solution

We have ( )( )! 2n !n u

2

n = and we simplify U n+1 to obtain

( )( )( )( )( ) ( )

( )! 22n! 1n ! 1n

! 1n2! 1nu

2

1n +++=+

+=+

= ( ) ( )( )( ) ( )! 2n 12n22n1n !n 1nn!

++++

= ( )( )( )

( )( )12n22n1n

! 2n!n 22

+++

= ( )( )( )12n1n21nu

2n ++

+

Thus ( )( )12n21n

uu

n

1n+

+=+ after simplification. We then find that

.41

n24

n11

lim24n

1nlimu

ulim

nnn

1nn

=

+

+=+

+= +

Here 141L

24

We apply the Ratio-Test to both series:

for =1n n

1 , we have

n1

1n1

limU

Ulim

nn

1nn

+= +

= 1n

nlimn +

=

n11

1limn +

= 1 .

No conclusion can be reached by the Ratio-Test. However, we know that =1n n

1

diverges.

For =1n 2n

1 , we have ( )2

2

nn

1nn

n11n

1

limu

ulim +=

+

= ( ) 11nnlim

2

2

n=

+

The Ratio-Test provides no information in this case also. But, we have seen that =1n 2n

1

is a convergent p-series with p = 2.

Thus, we have shown that when 1u

ulim

n

1nn

=+ , the series

=1nnu may either converge or

diverge. Activity 6 Do the following series converge or diverge?

(a) =1n n n!2

1 (b) =1n n

4

2n (c)

=

1n

n2en (d) =1n

3

n!n

25

4.4 STRATEGIES FOR SERIES Here is a general set of guidelines to help you determine the convergence of a series. Note that these are a general set of guidelines and because some series can have more than one test applied to them, we will get a different result depending on the path that we take through this set of guidelines. 1. With a quick glance does it look like the series terms do not converge to zero in

the limit, i.e. does 0 ulim nn

? If so, use the Divergence Test. Note that you should only do the divergence test if a quick glance suggests that the series terms may not converge to zero in the limit.

2. Is it a p-series, =1 n

pn1 or a geometric series

= 0 n nr a ? If so, then p-series converge

for p > 1 and a geometric series converges if 1 r < . 3. Does the series behave as a p-series or a geometric series for large values of n? If

so, use the comparison test. 4. Is the series a rational expression involving only polynomials or polynomials

under radicals (i.e. a fraction involving only polynomials or polynomials under radicals)? If so, try the Comparison Test or the Limit Comparison Test.

5. Does the series contain factorials or constants raised to powers involving n? If so, then the Ratio Test may work. Note that if the series term contains a factorial then very often, the most suitable test is the Ratio Test.

26

4.5 POWER SERIES When studying infinite series, we have given some tests to determine whether a given series converges or not. In this section, we study infinite polynomials called Power Series.

Definition

A power series is a series of the form

( ) ( ) ( ) ( )n

nnc x a c c x a c x a c x a

=

= + + + +0

0 1 22

33 ... ( )+ +c x an n ...

in which the center a and the coefficients c c c cn0 1 2, , ,..., ,... are constants.

Example 11

The geometric power series with center 0 is given by

n

n nx x x x x=

= + + + + + +0

2 31 ... ...

The power series converges to 11 x when x < 1.

Example 12

The power series ( ) ( ) ( )1 13

1 19

1 13

12 + + + +x x x

nn... ...

has center a = 1 and cnn

=

13

.

27

Expressing ( ) ( ) ( )1 13

1 19

1 13

12 + + + +x x x

nn... ... in the form ( )

n

nnx

=

0

13

1 , we find that for a fixed value x, the series is a geometric series.

Thus the series converges provided the common ratio r satisfies r < 1.

Here ( )r x= 13

1 , and hence there is convergence when

( ) 13

1x < 1 , that is, for -2 < x < 4 ,

and the sum is given by

( ) ( )1

1 13

1

33 1

32

= + = +x x x

Thus we can write

( ) ( ) ( )32

1 13

1 19

1 13

12+ = + + +

+x x x x

nn... ... for -2 < x < 4 .

We find that the function 32 + x can be expressed as a power series for <

28

4.6 TAYLOR SERIES AND MACLAURIN SERIES One of the most important theorems in Calculus is Taylors Theorem. This theorem provides an estimate for the error involved when a function ( )f x is approximated near the point x = a by the polynomial of degree n in (x - a) which best describes the behavior of the function ( )f x near that point. 4.6.1 TAYLORS THEOREM Let f have continuous derivatives up to, and including order n + 1 on some interval containing the point x a= . Then, if x is any other point in the interval, f(x) can be expressed in the form ( ) ( ) ( )f x P x R xn n= + +1 where

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )P x f a x a f a x a f a x an

f ann

n= + + + + '!

" ...!

2

2

and ( ) ( )( ) ( )R xx an

fnn

n+

++= +1

11

1 ! where a < < x

( )P xn is called the Taylor polynomial of order n generated by f at a and ( )R xn+1 is the remainder term. Example 13

Find the Taylor polynomial of order 3 generated by ( )f xx

= 1 about x = 1. Write down the remainder term.

29

Solution Here a = 1 and the Taylor polynomial of order 3 is given by

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )P x f x f x f x f3

2 331 1 1

12

11

31= + + + '

! !"

We compute ( ) ( ) ( )f f f1 1 1, , and ( )f 3 1 . We find that ( )f 1 1= and

( ) = = f xx

x12

2 ( )f ' 1 1=

( )f x x" = 2 3 ( )f " 1 2=

( )f x x3 46= ( )f 3 1 6=

( )f x x4 524= ( )f 4 1 24= We then have ( ) ( )( ) ( ) ( ) ( ) ( )P x x x x

3

2 3

1 1 11

22

13

6= + + + ! !

.

Thus, the cubic approximation to the function 1x

about x = 1 is given by

( ) ( ) ( ) ( )P x x x x = + 1 1 1 12 3 .

The remainder term is ( ) ( ) ( )R x x f4 4 414= ! where 1 < < x . Thus ( ) ( ) ( )R x x x4

4

5

4

5

124

24 1= = .

Thus we can write

( ) ( ) ( ) ( ) ( )f x x x x x= + + 1 1 1 1 12 34

5 .

30

Example 14

Find the Taylor polynomial of order 4 generated by ( )f x e x= at a = 0. Write down the remainder term.

Solution

( )f x ex= ( )f e0 10= =

( )f x e x' = ( )f e' 0 10= =

( )f x e x" = ( )f e" 0 10= =

( )f x ex3 = ( )f e3 00 1= =

( )f x ex4 = ( )f e4 00 1= = The Taylor polynomial of order 4 is

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )P x f x f x f x f x f42 3

34

40 0 00

20

03

00

40= + + + + ' "

! ! !

( ) ( ) ( ) ( )= + + + + = + + + +1 12

13

14

1 12 3 4

2 3 4 2 3 4

xx x x

xx x x

! ! ! ! ! !

The remainder term ( )R x5 is given by ( ) ( ) ( )R x x f5 5 505= ! where a < < x

Since ( )f x ex5 = , we obtain ( )R x x e5 55= ! . Thus we can write

e x x x x x ex = + + + + +12 3 4 5

2 3 4 5

! ! ! ! where 0 < < x.

31

Activity 7

Find the Taylor polynomial of order 4 generated by ( )f x x= cos at a = 3

.

The Taylor series corresponding to a = 0 is called a Maclaurin series.

32

4.6.2 TAYLORS FORMULA FOR FUNCTIONS OF TWO VARIABLES In this section, we extend Taylors formula introduced in Section 4.6, to functions of two variables. Taylors Formula for ( )f x y, at the point (a,b) Suppose ( )f x y, and its partial derivatives are continuous throughout an open rectangular region R centred at a point (a,b). Then throughout R

( ) ( ) ( ) ( )f a h b k f a b h fx

a b kfy

a b+ + = + + +, , , ,

( ) ( ) ( )

( )

12

2

13

3 3

22

2

22

2

2

33

32

3

22

3

23

3

3

!, , ,

!...

,

hf

xa b hk

fx y

a b kf

ya b

hf

xh k

fx y

hkf

x yk

fy

a b

+ +

+ + + + +

or for ( )x y, R

( ) ( ) ( ) ( ) ( ) ( )f x y f a b x a fx

a b y bfy

a b, , , ,= + + +

( ) ( ) ( )( ) ( ) ( ) ( )12

222

2

22

2

2!, , , ...x a

fx

a b x a y bf

x ya b y b

fy

a b + + +

33

Example 15 Use Taylors formula to find a quadratic polynomial that approximates ( )f x y x y, sinh sinh= near the origin. Solution The required quadratic polynomial

( ) ( ) ( ) ( ) ( ) ( )f x y f x fx

yfy

, , , , + + 0 0 0 0 0 0 0 0

( ) ( ) ( ) ( )+ + +

12

0 0 0 2 0 0 0 022

2

22

2

2!, , ,x

fx

xyf

x yy

fy

.

( )f x y, = sinh x sinh y ( )f 0 0, = sinh 0 sinh 0 = 0

fx

= cosh x sinh y ( )fx

0 0, = cosh 0 sinh 0 = 0

fy

= sinh x cosh y ( )fy

0 0, = sinh 0 cosh 0 = 0

2

2

fx

= sin x sinh y ( )2

2 0 0 0f

x, =

2 f

x y= cosh x cosh y ( )

2

0 0f

x y, = cosh 0 cosh 0 = 1

2

2

fy

x y= sinh cosh ( )2

2 0 0 0f

y, =

Thus

( ) ( ) ( ) [ ]f x y x y x xy y xy, . . ! . . . + + + + + 0 0 0 0 0 12 0 2 1 02 2 The quadratic polynomial is xy.

34

Example 16

Use Taylors formula to find a quadratic polynomial that approximates ( )ylnex +1 near the origin.

Solution

Here

( ) ( )y1lneyx,f x += ( ) ( ) 01ln101lne0,0f 0 ==+=

( )y1nlexf x +=

( ) 00,0xf =

y1e

yf x

+= ( ) 1

01e0,0

yf 0 =+=

( )y1lnex

f x2

2+=

( ) 0ln1e0,0x

f 02

2==

y1e

yxf x2

+= ( ) 10,0

yxf2 =

( )2x

2

2

y1e

yf

+=

( ) ( ) 101e0,0

yf

2

0

2

2=+

=

The required quadratic polynomial is

( ) ( ) ( ) ( ) ( )[ ]1.0y2xy.1.00x2!1.10y.00x0yx,f 22 +++++

( )2y2xy21y +

35

+ y xy y12

2

Activity 8

Use Taylors formula for ( )y,xf at the origin to find a quadratic polynomial approximation of f near the origin : (i) ( ) ycosey,xf x 32=

(ii) ( ) =y,xf cos ( )yx +

(iii) ( )yx

y,xf = 11

36

4.7 SUMMARY In this unit, you have learnt the following: 1. An infinite series is an expression of the form

u1 + u2 + + un + = =1n

nu .

The sum given by =

=N

nnN uS

1 gives a sequence of partial sums { }NS .

2. An infinite series is said to be convergent to the limit L if the sequence of partial

sums converges to the limit L. Otherwise, it is said to be divergent. 3. A geometric series is convergent if r < 1 and divergent if r 1 4. The nth Term Test for Divergence:

The series =1n

nu is divergent if for the nth term un , nn

ulim+ is non zero or does

not exist.

37

5. Comparison Test for Convergence

Let =1n

nu be a series such that Un > 0 for all n.

(i) Suppose 0 , where 0n is some positive integer. If =1n

nv

converges, then =1n

nu converges.

(ii) Suppose nn uv . If =1n

nv diverges, then =1n

nu diverges.

The series =1n pn

1 converges if p>1 and diverges if 1p

6. Limit Comparison Test

Suppose that 0u n > and 0>nv for all 0n > and that =+ nn

n vulim L.

(iv) If

38

7. Ratio Test

Let =1n

nu be a series with 0u n > and suppose Luulim

n

1nn

=++ , then

(i) The series converges if L < 1.

(ii) The series diverges if L > 1.

(iii) The series may converge or diverge if L = 1.

8. Taylors Theorem Let f have continuous derivatives up to, and including order n+1 on some interval containing the point x = a. Then, if x is any other point in the interval, ( ) ( ) ( )xRxPxf 1nn ++= where the Taylor polynomial is given by

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )af!nax...af

!2axafaxafxP n

n2

n++++=

and the remainder term is given by

( ) ( )( ) 1n1n

1n f!1naxxR +

++ +

= () for

39

4.8 ANSWERS TO ACTIVITIES

Activity 1

(a) Note that

+

=

+ n 1ln n n

n

ne lim

n 1 lim ,

by the continuity of the exponential function,

e e lim n 1ln n lim

n 1ln n

n

n

+

+

= .

Let n1 x = , we then get n as 0 x and

Therefore,

( ) x 1ln x

limn 1ln n lim

0 +

+

= xn ee .

Applying LHopitals rule, we hence obtain ( )

x 1 lim

x x 1ln lim

xx e e ++ = 00 = e

(b) e2

40

Activity 2 (a) Write the term of the series as partial fractions :

2 n 1 -

1 n 1

2 n 3 n1

2 ++=++

1 n + We write the terms of the general partial sum for this series using the partial fraction form to get

This is a telecosping sum and the partial sum is given by :

2n 1 - 1 Sn +=

1 2 n

1 - 1 lim Slim n

n n

=

+=

The sequence of partial sums is convergent and so the series is converges to 1.

(b) same procedure as in (a). convergent and limit is 125 .

(c) =

=

=

++=

++0 n

n 5-n

0 n 2

0 n

n 5-n2 e 7 - 3 n 4 n

1 e 7 - 3 n 4 n

1

and this is the sum of two convergent sequences. Hence it is a convergent sequence. Activity 3

(a) 21-

n 2 10n - n 4 lim

3

32

n =+ . The limit of the series terms is not zero and so by nth-term

test for divergence the series diverges.

(b)(i) I When

+

+== n as 41

3 n 21 n u 1,

2

nx . Therefore,

(ii) In that case, n as 0 nu and we can not conclude anything about the convergence/divergence of the series. This is neither a sufficient condition for convergence nor does it indicate that the series diverges. We need to revert to other tests.

(c)6 n 7

n 5 lim2 n +

= 7

5

n6 7

5 lim

2 n

=+

.

Hence the series diverges by the nth term test for divergence.

+++

+++

+

+

=2 n

1 1 n

1 1 n

1 n1

41 -

31

31 -

21

21 -

11 Sn K

n as 0 nu

41

(d) 1 7lim 7lim n1

n n

n ==

Hence the series diverges by the nth term test for divergence. Activity 4

(a) += n allfor , 21

121 u 0

nnn. Since

=1n n21 converges (it is a geometric

series with common ratio r = 21 ), it follows, by the comparison test

= +1 n n 1 21

converges

(b) For all integers n > 0, 323 n

1 5 n 4 n

1

42

(d) For large n, sin 2n

1 is approximately equal to 2n

1 as 2n

1 is small.

We take 2n n

1sinu = and 2n n

1v = .

Then

2

2

nn

nn

n1n1sin

limvu

lim =

= x

xsinlim0x by putting 2n

1x =

= 1

coslim0

xx by lHpitals Rule

= cos 0

11= (0 < L < )

The series =1n 2n

1 converges (this is a p-series with p = 2 > 1).

Therefore =1n 2n

1sin also converges.

Activity 6 (a) Convergent (b) Convergent (c) Divergent

(d) Convergent (by the comparison test) as sin2

3 3

1nn n

since sin2 n 1 . As 131 nn=

converges,

sin23

1

nnn=

converges.

43

Activity 7 12

32 3

14 3

312 3

148 3

2 3 4

+

+

x x x x

Activity 8

(i) 1 2 2 92

22

+ + x x y

(ii) 1 12

12

2 2 x xy y (iii) 1 22 2+ + + + +x y x xy y

/ColorImageDict > /JPEG2000ColorACSImageDict > /JPEG2000ColorImageDict > /AntiAliasGrayImages false /DownsampleGrayImages true /GrayImageDownsampleType /Bicubic /GrayImageResolution 300 /GrayImageDepth -1 /GrayImageDownsampleThreshold 1.50000 /EncodeGrayImages true /GrayImageFilter /DCTEncode /AutoFilterGrayImages true /GrayImageAutoFilterStrategy /JPEG /GrayACSImageDict > /GrayImageDict > /JPEG2000GrayACSImageDict > /JPEG2000GrayImageDict > /AntiAliasMonoImages false /DownsampleMonoImages true /MonoImageDownsampleType /Bicubic /MonoImageResolution 1200 /MonoImageDepth -1 /MonoImageDownsampleThreshold 1.50000 /EncodeMonoImages true /MonoImageFilter /CCITTFaxEncode /MonoImageDict > /AllowPSXObjects false /PDFX1aCheck false /PDFX3Check false /PDFXCompliantPDFOnly false /PDFXNoTrimBoxError true /PDFXTrimBoxToMediaBoxOffset [ 0.00000 0.00000 0.00000 0.00000 ] /PDFXSetBleedBoxToMediaBox true /PDFXBleedBoxToTrimBoxOffset [ 0.00000 0.00000 0.00000 0.00000 ] /PDFXOutputIntentProfile () /PDFXOutputCondition () /PDFXRegistryName (http://www.color.org) /PDFXTrapped /Unknown

/Description >>> setdistillerparams> setpagedevice