CHAPTER 5 INFINITE SEQUENCES AND SERIES SERIES

CHAPTER 5 INFINITE SEQUENCES AND SERIES... Taylor and Maclaurin Series 5.8.. Using Series to solve Differential Equations 5.11... Properties of Convergent Series... We may say that the

CHAPTER 5 INFINITE SEQUENCES

AND SERIES

CONTENTS

5.1 Sequences

5.2 Series

5.3 The Integral and Comparison Test

5.4 Other Convergence Test

5.5 Power Series

5.6 Representations of Functions as Power Series 5.7 Taylor and Maclaurin Series

5.8 The Binomial Series

5.9 Applications of Taylor Polynomial

5.10 Using Series to solve Differential Equations 5.11 Fourier Series

5.2 SERIES

5.2.1 The Sum of a Series

5.2.2 Geometric Series

5.2.3 The Test for Divergence

5.2.4 Properties of Convergent Series

5.2.1 The Sum of a Series

 We can add the terms of a sequence {a n } and get an expression of the form:

a1+ a2+ a3+ …+ a n + … which is called a series and denoted by

Example We can try to add the terms of the series

1+2+3+…+n+…

and get the cumulative sums

1, 3, 6, 10, …,

The nth sum n(n+1)/2 becomes very large as n increases

we get the cumulative sums

 On the other hand if we add the terms of the series

14

12

,

,8

7,4

3,2

1

n



We may say that the sum of this infinite series is 1 in the following sense:

otherwise, it is divergent This number s is called

the sum of the series

Let s n denote the nth partial sum

1

s a

a a

So now we can write

1 2

1 4

1 2

1 2

Let's consider the geometric series

ar ar

a

n

n n





5.2.2 The Geometric Series

Let's consider the geometric series

If r = 1, then the nth sum s n =na   as n   Hence the geometric series is divergent

ar ar

a

n

n n





n n

n

n n

ar ar

ar ar

rs

ar ar

ar a

1 2

If –1<r<1, then we know that r n  0 as n   So

r

a r

r

a r

a r

(limlim

Thus the geometric series is convergent; its sum is

r

a s





1

If r  –1 or r >1, then {r n} is divergent so the limit

of s n as n   does not exist Thus the geometric series is divergent

r

r

a s

(

s n -rs n = a - ar n

The geometric series

is convergent if r < 1 and its sum is

a ar

r

a ar

if r  1, the geometric series is divergent

Example Find the sum of the geometric series

10

5

The geometric series

is convergent if r < 1 and its sum is

a ar

r

a ar

if r  1, the geometric series is divergent

Example Find the sum of the geometric series

(1

55

3

5 3

2 27

40 9

20 3

Example Is the series  convergent?



 1

1 2

3

2

n

n n

Example Is the series  convergent?



 1

1 2

3

2

n

n n

Solution We rewrite the nth term of the series in

1

1 2

3

443

43

n n

n

Therefore the series is a geometric series with a=4 and r = 4/3 > 1 so it is divergent

Example Write 2.317  2.3171717 as a fraction

Example Write as a fraction

Solution We may write

1710

173

.2

3171717

2

Apart from the first term 2.3, the rest is the sum of

a geometric series with a=17/103 and r = 1/102 < 1

so it is convergent, and we have

3171717

217

3

495

1147990

1710

23

11000

173

21

110

173

.217

3.2

100

99 10

1 3

Example Find the sum of the series

1where

Example Find the sum of the series

Solution This is a geometric series with a=1 and

r = x Since r < 1, the series is convergent; its sum is

1where

Example Find the sum of the series



1 ( 1)

1

n n n

11

1

11

4

13

13

12

12

11

1

11

)1(

1

1 1

n

i i

i i s

n i

n i

n



10

11

1lim

1lim

1(

Example Determine whether the series is

convergent or divergent If the series is convergent, find its sum

Theorem 1 If the series is convergent,

lim)

(lim

n n

n n

n

1lim 

Note The converse of Theorem 1 is not true as

shown by the following example

12

11

1

1

Example Show that the series

is divergent

The Test for Divergence If does not exist

or , then the series is divergent

n n

n

Example Show that the series

is divergent

Solution We have

05

14

5

1lim

45

lim

2 2

n

n n

The Test for Divergence If does not exist

or , then the series is divergent

n n

n

Therefore the series is divergent by the Test for

Divergence

Example Show that the series is divergent

a. ∞𝑛=1 3𝑛+12𝑛

b. ∞𝑛=1 sin 𝑛 (or ∞𝑛=1 sin 𝑛 𝑥 where 𝑥 ≠ 𝑘𝜋 )

c. 1 + 13 + 15 + 17 + ⋯ … + 2𝑛−11 + ⋯

Theorem If are convergent series

and c is a constant, then the following series are

1

1 1

1

1 1

) (

) (

n

n n

n n

n n

n

n n

n n

n n

n

n n

n

b a

b a

b a

b a

a c

n b a

Example Find the sum of the series 

1(

3

n

n

n n

Example Find the sum of the series

Solution1. We have seen that the series

is convergent and its sum is 1

On the other hand the series is a geometric series with

1(

3

n

n

n n

1

2 1 2 1

13

2

1)

1(

13

2

1)

1(

3

1 1

n

n

n n n

n

Example Find the sum of the series 31𝑛 + 2

(3𝑛 − 2)(3𝑛 + 1)

∞

𝑛=1