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CHAPTER 5 : SERIES

5.1 Series

5.2 The Sum of a Series

5.2.1 Sum of Power of ‘n’ Positive Integers

5.2.2 Sum of Series of Partial Fraction

5.2.3 Difference Method

5.3 Test of convergence

5.3.1 Divergence Test

5.3.2 Integral Test

5.3.3 Ratio Test

5.4 Power Series

5.5 Taylor and the Maclaurin Series

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Review:

Sequence

What is a sequence? It is a set of numbers which are written

in some particular order

1 2, , , .nu u u

We sometimes write 1u for the first term of the sequence,

2u for the second term and so on. We write the nth term as

nu .

Examples:

1, 3, 5, 9. – finite sequence

1, 2, 3, 4, 5, …, n – finite sequence

1, 1, 2, 3, 5, 8, … - infinite sequence

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5.1 Series

Definition:

The sum of the terms in the sequence is called a series.

For example, suppose we have the sequence

1 2, , , .nu u u

The series we obtain from this is

1 2 .nu u u

and we write nS for the sum of these n terms.

For example, let us consider the sequence of numbers

1, 2, 3, 4, 5, 6, . . . , n .

Then,

1

2

3

1

1 2 3

1 2 3 6

S

S

S

The difference between the sum of two consecutive partial

terms, 1n nS S , is the nth term of the series.

i.e 1n n nu S S

If the sum of the terms ends after a few terms, then the

series is called finite series.

If the sum of the series does not end, then the series is

called infinite series.

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Example 1:

Find the 4th term and 5th term of the sequence 1, 4, 7, ….

Hence, find S4 and S5 of the series 1, 4, 7, ….

Example 2:

The sum of the first n terms of the series is given by

21(5 11 )

4nS n n

a) Find the first three terms, and

b) The n-th term of the series.

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Summation Notation, ∑ .

∑ (read as sigma) is used to represent the sum of the series.

In general,

1 2

1

.n

n n r

r

S u u u u

(finite)

1 2 3

1

.i

i

S u u u u

(infinite)

Example 3:

Find the r-th term of the following series. Hence, express

the series using ∑ notation.

a) 2 3 4 ..., to 10 terms.

b) 3 9 27 ...,until 30 terms.

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Before we proceed to the next sub-topic, let us review two

important sequences/progressions, i.e.

1. Geometric Sequence

2. Arithmetic Sequence

Geometric Sequence

- GS is a sequence where each new term after the first is

obtained by multiplying the preceding term by a

constant r, called the common ratio.

- If the first term of the sequence is a, then the GS is 2 3, , , ,...a ar ar ar

where the n-th term is 1.nar

- E.g. 2, 6, 18, 54, .. ( 2, 3)a r

1, -2, 4, -8 ( 1, 2)a r

Arithmetic Sequence

- AS is a sequence where each new term after the first is

obtained by adding a constant d, called the common

difference to the preceding term.

- If the term of the sequence is a, then the AS is , , 2 , 3 ,...a a d a d a d

where the n-th term is ( 1) .a n d

- E.g. 8, 5, 2, -1, -4, .. ( 8, 3)a d

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5.2 Sum of Series

5.2.1 Sum of Power of ‘n’ Positive Integers

1

11 2 3

2

n

r

n nr n

2 2 2 2 2

1

11 2 3 1 2 1

6

n

r

r n n n n

2

3 3 3 3 3

1

11 2 3

2

n

r

n nr n

Example 4:

Evaluate

202

1r

r

and

253

1r

r

.

Example 5:

Evaluate 10

2

1

2 1r

r

.

Example 6:

Find the sum for each of the following series:

(a) 22 2 22 4 6 2n

(b) 1 3 4 5 7 7 to 30 terms

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5.2.2 Sum of Series of Partial Fraction

In this section we shall discuss terms with partial fractions

such as 1 1 1

...2 3 3 4 4 5

We are not able to calculate the sum of the series by using

the available formula (so far), but with the help of partial

fraction method, we can solve the problem.

Example 7:

Find the sum of the first n terms of the series 1 1 1

...2 3 3 4 4 5

The above problem requires quite a long solution. However,

in the next sub-topic, we will see a different approach to

solve the same problem. We called the approach, a

difference method.

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5.2.3 Difference method

Let ( )f x be a function of x and the r-th term of the series

1

n

rr

u is of the form ( ) ( 1)ru f r f r , then

1 1

( ) ( 1)

(1) (0) (2) (1) (3) (2)

(4) (3) ... ( ) ( 1)

(0) ( )

( ) (0).

n n

rr r

u f r f r

f f f f f f

f f f n f n

f f n

f n f

To conclude,

If 1ru f r f r , then 1

0 .n

r

r

u f n f

Or equivalently

If 1rku f r f r , then 1

10 .

n

r

r

u f n fk

where k is a constant.

Note: If we fail to express ru into this form, i.e.

1f r f r , then this method cannot be used.

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Example 8:

Express the r-th term of the series

1 2 2 3 3 4 ... ( 1) ...r r as the difference of two

functions of r and r – 1. Hence find the sum of the first n

terms of the series.

Solution:

Step 1: Find the general form of the r-th term:

Step 2: Form another sequence ( )f r by adding one more

factor to the end of the general term ur :

Step 3: Find ( 1)f r :

Step 4: Form the difference:

( ) ( 1)f r f r

Step 5: Find the sum :

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Tips:

(A) If the general term, ru , of the series is in "product" form,

you can add one more factor to the end of the general term ru , so

as to form a sequence ( )f r and then apply the difference

method.

(B) If the general term, ru , is in "quotient" form, you can

remove one more factor at the end of the general term ru , so as

to form a sequence ( )f r and then apply the difference method.

Example 9:

By using the difference method, find the sum of the first n

terms of the series 1 1 1 1

... .2 3 3 4 4 5 ( 1)( 2)n n

Example (10):

Use the difference method; find the sum of the series

1

2

2 3

n

r r r .

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5.3 Test of Convergence

5.3.1 Divergence Test

If 1

rr

a

converges, then 0lim r

r

a

. Equivalently, if

0lim rr

a

, or lim rr

a

does not exist, then the series is

diverges.

Example (11):

Show that the series

2

21 5 4n

n

n

diverges.

Example 12:

Use Divergence Test to determine whether 1 lnr

r

r

diverges

or not.

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5.3.2 The Integral Test

Suppose f is a continuous, positive, decreasing function on

1, and let ra f r . Then the series

1

r

r

a

is convergent

if and only if the improper integral 1

f x dx

is

convergent. In other words

(a) If 1

f x dx

is convergent, then

1

r

r

a

is

convergent.

(b) If 1

f x dx

is divergent, then

1

r

r

a

is divergent.

Note: Use this test when f x is easy to integrate.

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Example 13:

Use the Integral Test to determine whether the following

series converges or diverges.

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5.3.3 Ratio Test

Let 1

rr

a

be an infinite series with positive terms and let

1 .limr

r r

a

a

a) If 0 1 , the series converges.

b) If 1 , or , the series diverges.

c) If 1 , the test is inconclusive.

Example 14:

Use the Ratio Test to determine whether the following

series converges or diverges.

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5.4 Power Series

Definition

A power series about 0x is a series of the form

2 3

0 1 2 3

0

   n n

n n

n

a x a a x a x a x a x

A power series about x a is a series of the form

2

0 1 2

0

( )   ( )   ( ) ( )n n

n n

n

a x a a a x a a x a a x a

in which the center a and the coefficients

0 1 2,, , , ,na a a a are constants.

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5.4.1 Expansion of Exponent Function

The power series of the exponent function can be written as

2 3 41 1 11

2! 3! 4!

xe x x x x

The expansion is true for all values of .x In general,

0

1

!.x n

n

e xn

Example (15):

Given

2 3 41 1 1 11

2! 3! 4! !

x ne x x x x xn

Write down the first five terms of the expansion of the

following functions

(a) 2xe

(b) 1xe

Example (16):

Write down the first five terms on the expansion of the

function, 2

1 xx e in the form of power series.

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5.4.2 Expansion of Logarithmic Function

The expansion of logarithmic function can be written as

2 3 4 5

6 7

1 1 1 1ln(1 x)

2 3 4 5

1 1

6 7

x x x x x

x x

The series converges for 1 1x . Thus the series

ln 1 x is valid for 1 1x .

By assuming x with x , we obtain

2 3 4 5

6 7

1 1 1 1ln(1 x)

2 3 4 5

1 1

6 7

x x x x x

x x

Thus, this result is true for 1 1x or 1 1.x

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Example (17):

Write down the first five terms of the expansion of the

following functions

(a) ln 1 3x

(b) 23ln 1 2 1 3x x

Example (18):

Find the first four terms of the expansion of the function,

2 3

1 ln 1 2 .x x

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5.4.3 Expansion of Trigonometric Function

The power series for trigonometric functions can be written

as 3 5 7 9

sin      3! 5! 7! 9!

x x x xx x

2 4 6 8

cos     1 2! 4! 6! 8!

x x x xx

Both series are valid for all values of .x

Example (19):

Given 2 4 6 8

cos     1 2! 4! 6! 8!

x x x xx

Find the expansion of cos 2x and cos 3x . Hence, by

using an appropriate trigonometric identity find the first

four terms of the expansion of the following functions:

(a) 2sin x

(b) 3cos x

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5.5 Taylor and the Maclaurin Series

Example 20:

Obtain the Taylor series for 2( ) 3 6 5f x x x around the point

1.x Ans: 2 + 3(x-1)2

Example 21: Obtain Maclaurin series expansion for the first four terms of

xe and five terms of sin x . Hence, deduct that Maclaurin series for

sinxe x is given by 2 3 51 1...

3 30x x x x

Example 22:

Use Taylor’s theorem to obtain a series expansion of first five

derivatives for cos3

x

. Hence find 0cos62 correct to 4 dcp.

Ans: 0.4695

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Example 23:

If lncosy x , show that 22

21 0

d y dy

dx dx

Hence, by differentiating the above expression several times,

obtain the Maclaurin’s series of lncosy x in the ascending

power of x up to the term containing 4x .

Solution:

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Finding Limits with Taylor Series and Maclaurin Series.

Example 24:

Find 20

1lim

x

x

e x

x

.

Ans: ½

Example 25:

Evaluate 2

40

2cos 2lim

3x

x x

x

.

Ans: 1/36

Evaluating Definite Integrals with Taylor Series and Maclaurin

Series.

Example 26:

Use the first 6 terms of the Maclaurin series to approximate the

following definite integral.

a) 21

0

xe dx

b) 1

3

0

cos( )x x dx

Ans: 0.747, 0.440