INFINITE SERIES Sequence - · PDF fileINFINITE SERIES Sequence: If a set of real numbers occur...

INFINITE SERIES

Sequence:

If a set of real numbers occur according to some definite rule, then it is called

a sequence denoted by * + * + if n is finite

Or * + * + if n is infinite.

Series:

is called a series and is denoted by ∑

Infinite Series:

If the number of terms in the series is infinitely large, then it is called infinite series and is

denoted by ∑ and the sum of its first n terms be denoted

by ∑ .

Convergence:

An infinite series ∑ is said to be convergent if , a definite unique number.

Example:

.

/

.

/ , finite.

Therefore given series is convergent.

Divergence:

tends to either then the infinite series ∑ is said to be divergent.

Example: ∑

( )

Therefore ∑ is divergent.

Oscillatory Series:

If tends to more than one limit either finite or infinite, then the infinite series ∑

is said to be oscillatory series.

Example:1. ∑ :

{

Therefore series is oscillatory.

2. ∑ ( ) ( )

( )

{

Properties of infinite series:

1. The convergence or divergence of an infinite series remains unaltered on

multiplication of each term by .

2. The convergence or divergence of an infinite series remains unaltered by addition or

removal of a finite number of its terms.

Positive term series:

An infinite series in which all the terms after some particular term are positive is called a

positive term series.

Geometric Series test:

The series ∑

a. Converges if | |

b. Diverges if

c. Oscillates finitely if and oscillates infinitely if

Proof:

Let be the partial sum of ∑ .

Case 1: | | i.e.

Therefore the series is convergent.

Case 2i: i.e.

Therefore the series is divergent.

Case 2ii: ,

. Therefore the series is divergent.

Case 3i: i.e.

( )

{

Therefore the series is oscillatory.

Case 3ii:

i.e.

{

Therefore the series is oscillatory.

Note: If a series in which all the terms are positive is convergent, the series remains

convergent even when some or all of its terms are negative.

Integral Test:

A positive term series ( ) ( ) ( ) Where ( ) decreases as n

increases, converges or diverges according as the integral ∫ ( )

is finite or infinite.

p-series or Harmonic series test:

A positive term series ∑ ∑

is

i) Convergent if

ii) Divergent if

Proof:

Let ( )

∫ ( ) ∫

[

]

{

{

When , ∫ ( ) ∫

, -

Thus ∑

converges if and diverges if .

Theorem:

Let ∑ be a positive term series. If ∑ is convergent then .

Proof:

If ∑ is convergent then .

( ) ( )

Note:

Converse need not be always true. i.e. Even if , then ∑ need not be

convergent.

Example 1: ∑

.

∑

is divergent by integral test. But

Hence is a necessary condition but not a sufficient condition for convergence

of ∑ .

Example 2

Test the series for convergence, ∑

Solution: Consider ∫

, ( )-

Therefore ∑ is divergent by Integral test.

Example 2


Solution: Let . Then

∫ ∫

[

]

Therefore ∑ is convergent.

Comparison test:

1. Let ∑ and ∑ be two positive term series. If

a. ∑ is convergent

b.

Then ∑ is also convergent.

That is if a larger series converges then smaller also converge.

2. Let ∑ and ∑ be two positive term series. If

c. ∑ is divergent

d.

Then ∑ is also divergent.

That is if a smaller series diverges then larger also diverges.

Example 2


Solution:

Let

and

n n

log n n

1 1

log n n

u v

But ∑ ∑

is a p-series with .

Therefore ∑ is divergent.

By comparison test ∑ is also divergent.

Example 2


Solution:

Let

and

n n

n n

n n

2 2 1

1 1

2 2 1

v u

But ∑ ∑

is a geometric series with

.

Therefore ∑ is convergent.

By comparision test ∑ is also convergent.

Another form of comparison test is

Limit test

Statement: If ∑ and ∑ be two positive term series such that

( ).

Then ∑ and ∑ behave alike.

That is if ∑ converges then ∑ also converge.

If ∑ diverges then ∑ also diverge.

Examples 3.

Test the series for convergence,

( )( )

Choose

then

But ∑ ∑

with .

Therefore ∑ is convergent. By limit test ∑ is also convergent.

Examples 4.

Test the series for convergence, ∑ (√ )

Solution: (√ )(√ )

(√ )

√

(√ )

Let ∑ ∑

( )

But ∑ is divergent. By limit test ∑ is also divergent.

Examples 5.

Test the series for convergence, ∑ √

Solution:

13 3

n

3 3 2 2

3 3

2 2

3 313 3

n 2 13 3 23 3

2 13 3

2

3 3

u n 1 n

a b (a b)(a ab b )

a ba b

a ab b

n 1 nu n 1 n

n 1 (n 1) n n

1

1 1n 1 1 1

n n

Let ∑ ∑

with .

But ∑ is convergent. By limit test ∑ is also convergent.

Example 6.

Test the series for convergence, Solve √

√

√

Solution:

n 3 3

3

3

1 1n 1

n nn 1 1u

n 2 1 2 1n 1

n n

Let ∑ ∑

⁄ with

.


Example 7


Solution:

We know that

Let ∑ ∑

. Then


Example 8


.

/

Solution:

.

/

[

]

[

]

Let ∑ ∑

. Then


Exercises

Test for convergence of the series

1. ∑

2.

…… …

3.

…….. …

4. ∑√

5. ∑

( )

6.

…… …

INFINITE SERIES

D’Alembert’s Ratio Test: If ∑ is a series of positive terms, and

nlim

n

n

u

u 1

( )

then the series is convergent if , is divergent if and the test fails if .

If the test fails, one should apply comparison test or the Raabe’s test, as given below:

Raabe’s Test: If ∑ is a series of positive terms, and

.

/ ( ) then the series is convergent if , is divergent if

and the test fails if .

Remark: Ratio test can be applied when (i) does not have the form ⁄

(ii) nth

term has etc.

(iii) nth

term has ( ) ( ) ect.

(iv) the number of factors in numerator and denominator increase steadily, ex: (

)

Example : Test for convergence the series

1 + !2

2 2

+ !3

32

+ !4

4 2

+ ….

>> The given series is of the form !1

12

+ !2

2 2

+ !3

32

+ !4

3 + … whose n

th term is un =

!

2

n

n.

Therefore un+1 = !1

)1( 2

n

n

n

n

u

u 1 = !1

)1( 2

n

n2

!

n

n =

2

2)1(

n

n .

)!)(1(

!

nn

n

=

2

1

n

n

Therefore n

lim n

n

u

u 1 = n

lim

2

1

n

n =

nlim

2

11

nn = 0 < 1

Therefore by ratio test, un is convergent.

Example : Discuss the nature of the series

2.1

x +

3.2

2x +

4.3

3x + ….

>> un = )1( nn

x n

Therefore un+1 = )11)(1(

1

nn

x n

= )2)(1(

1

nn

x n

Now n

n

u

u 1 = )2)(1(

1

nn

x n

. nx

nn )1( =

2n

n x

Therefore n

limn

n

u

u 1 = n

lim 2n

nx =

nlim

)/21(

1

nx = x

Therefore by D’Alembert’s ratio test un is

1 x if divergent

1 x if convergent

And the test fails if x = 1

But when x = 1, un = )1(

1

nn

n

= )1(

1

nn =

nn 2

1

un is of order 1/n2 (p = 2 > 1) and hence un is convergent (when x = 1). Hence we

conclude that un is convergent x 1 and divergent if x > 1

Example : Find the nature of series 1 + 2

x +

5

2x +

10

3x + ….

>> Omitting the first term, the given series can be written in the form

112

1

x +

122

2

x +

132

3

x + … so that un =

12 n

x n

Therefore un+1 = 222

1

nn

x n

. 22

12

2

nn

nx =

nlim

)/2/21(

)/11(22

22

nnn

nn

.x

That is, n

lim n

n

u

u 1 = x

Hence by ratio test un is

1 x if divergent

1 x if convergent

and the test fails if x = 1.

But when x = 1, un = 1

12 n

n

= 1

12 n

is of order 2

1

n (p = 2 > 1)

Therefore un is convergent if x 1 and divergent if x > 1.

Example: Find the nature of the series 12

1 +

23

2x +

34

4x + …

>> omitting the first term, the general term of the series is given by un = 1)2(

2

nn

x n

Therefore un+1 = 1)1()21(

)1(2

nn

nx =

2)3(

22

nn

x n

n

n

u

u 1 = 2)3(

22

nn

x n

nx

nn2

1)2(

= 3

2

n

n

2

2

1x

n

n

=

)3(

)1)(2(

n

nn x

2

nlim

n

n

u

u 1 = n

lim )/31(

)/11()/21(

nn

nnnn

. x

2 = x

2

Hence by ratio test un is

1 xif divergent

1 xif

2

2convergent

and the fails if x2 = 1.

When x2 = 1, un =

1)2(

)1(

nn

n

= 1)2(

1

nn

un is of order 1/n3/2

(p = 3/2 > 1) and hence un is convergent.

Therefore un is convergent if x2 1 and divergent if x

2 > 1.

Example : Discus the convergence of the series

x + 3.2

3x +

4.2

3 +

5

5x +

6.4.2

5.3 .

7

7x + … (x > 0)

>> We shall write the given series in the form

x + 2

1.

3

3x +

4.2

3.1.

5

5x +

6.4.2

5.3.1.

7

7x + ….

Now, omitting the first term we have

un = n

n

2...6.4.2

)12...(5.3.1 .

12

12

n

x n

un+1 = )1(2...6.4.2

]1)1(2...[5.3.1

n

n.

1)1(2

1)1(2

n

x n

That is, un+1 = )12....(6.4.2

)12...(5.3.1

n

n .

32

32

n

x n

That is, un+1 = )22)(2....(6.4.2

)12)(12...(5.3.1

nn

nn .

32

32

n

x n

Therefore n

n

u

u 1 = )22)(2....(6.4.2

)12)(12...(5.3.1

nn

nn.

32

32

n

x n

)12...(5.3.1

2...6.4.2

n

n.

12

12

nx

n

That is, n

n

u

u 1 = )32)(22(

)12)(12( 2

nn

xnn

Therefore n

lim n

n

u

u 1 = n

lim )/32()/22(

)/12()/12( 2

nnnn

xnnnn

= x

2

Hence by ratio test, un is

1 xif divergent

1 xif

2

2convergent

And the test fails if x2 = 1

When x2 = 1,

n

n

u

u 1 = )32)(22(

)12)(12(

nn

nn and we shall apply Raabe’s test.

nlim n

11n

n

u

u =

nlim n

1

)12(

)32)(22(2n

nn

= n

lim n

2

22

)12(

)144()6104(

n

nnnn

= n

lim n

2)12(

56

n

n =

nlim

22

2

)/12(

)/56(

nn

nn

4

6 =

2

3 > 1

Therefore un is convergent (when x2 = 1) by Rabbe’s test.

Hence we conclude that, un is convergent if x2 1 and divergent if x

2 > 1.

Example : Examine the convergence of

1 + 5

2x +

9

6 x

2 +

17

14x

3 + … +

12

221

1

n

n

xn + ….

>> un = 12

221

1

n

n

xn.

Therefore un+1 = 12

222

2

n

n

xn+1

n

n

u

u 1 = 12

222

2

n

n

xn+1

22

121

1

n

n

. nx

1

n

n

u

u 1 = )2/11(2

)2/21(222

22

nn

nn

.x. )2/21(2

)2/11(211

11

nn

nn

= )2/11(

)2/11(2

1

n

n

.x . )2/11(

)2/11( 1

n

n

Therefore n

lim n

n

u

u 1 = )01(

)01(

. x .

)01(

)01(

= x.

Therefore by ratio test un is

1 x if divergent

1 x if convergent and the test fails if x = 1.

When x = 1, un = 12

221

1

n

n

Therefore n

lim un = n

lim)2/11(2

)2/11(211

1

nn

nn

= 1

Since n

lim un = 1 0, un is divergent (when x = 1)

Hence un is convergent if x < 1 and divergent if x 1.

Example : test for convergence of the infinite series

1 + 22

!2 +

33

!3 +

44

!4 + …

>> the first term of the given series can be written as 1!/11 so that we have,

un = nn

n! and un+1 =

1)1(

)!1(

nn

n =

1)1(

)!)(1(

nn

nn =

nn

n

)1(

!

Therefore n

n

u

u 1 = nn

n

)1(

!

.

!n

n n

= n

n

n

n

)1( =

nn

n

nn

n

)/11(

nlim

n

n

u

u 1 = n

lim nn)/11(

1

=

e

1 < 1

Hence by ratio test un is convergent.

Cauchy’s Root Test: If ∑ is a series of positive terms, and

( ) ⁄ ( ),

then, the series converges if , diverges if and fails if

Remark: Root test is useful when the terms of the series are of the form , ( )- ( ).

We can note : (i) ⁄

(ii) ( ⁄ ) ⁄

(iii) ) ( ⁄ ) ⁄

Example : Test for convergence

1

2/3

11

n

n

n

>> un =

2/3

11

n

n

Therefore (un)1/n

=

nn

n

/12/3

11

=

2/1

11

n

n

=

n

n

11

nlim (un)

1/n =

nlim

n

n

11

= n

lim n

n

11

1 =

e

1 < 1.

Therefore as n , n also

Therefore by Cauchy’s root test, un is convergent.


1

2

31

n

n

n

>> un =

2

31

n

n

Therefore (un)1/n

=

nn

n

/12

31

=

n

n

31

nlim (un)

1/n =

nlim

n

4

31 = e

-3.

Therefore n

lim

n

n

x

1 = e

x

That is, n

lim (un)1/n

= 3

1

e < 1, therefore e = 2.7

Hence by Cauchy’s root test, un is convergent.

Example : Find the nature of the series

2/3

1

11

n

n n

>> un =

2/3

11

n

n

Therefore (un)1/n

=

nn

n

/12/3

11

=

2/1

11

n

n

=

n

n

11

nlim (un)

1/n =

nlim

n

n

11

= n

lim n

n

11

1 =

e

1 < 1, since as n , n also

Therefore by Cauchy’s root rest, un is convergent.


1

2

31

n

n

n

>> un =

2

31

n

n

Therefore (un)1/n

=

nn

n

/12

31

=

n

n

31

nlim (un)

1/n =

nlim

n

n

31 = e

-3, since

nlim

n

n

x

1 = e

x

That is, n

lim (un)1/n

= 3

1

e < 1, since e = 2.7.

Hence by Cauchy’s root test, un is convergent.

ALTERNATING SERIES

A series in which the terms are alternatively positive or negative is called an alternating

series.

i.e., 1

1 2 3 4

1

... 1n

n

n

u u u u u

LEBINITZ’S SERIES

An alternating series 1

1 2 3 4

1

... 1n

n

n

u u u u u

converges if

(i) each term is numerically less than its preceding term

(ii) lim 0nn

u

Note: If lim 0nn

u

then the given series is oscillatory.

Q Test the convergence of 6

1 -

13

1 +

20

1 -

27

1 + …

Solution: Here un = 17

1

n

then un+1 = 1)1(7

1

n =

67

1

n

therefore, un – un+1 = 17

1

n -

67

1

n

= )67)(17(

)17()67(

nn

nn =

)67)(17(

7

nn > 0

That is, un – un+1 > 0, un > un+1

Also, n

lim un = n

lim 17

1

n =

nlim

n

1

)/17(

1

n = 0

Therefore by Leibnitz test the given alternating series is convergent .

Q Find the nature of the series

2log

11 -

3log

11 +

4log

11 -

5log

11 + …

Solution: Here un = 1 - )1log(

1

n then un+1 = 1 -

)2log(

1

n

Therefore, un – un+1 = )2log(

1

n -

)1log(

1

n

= )1log()2log(

)2log()1log(

nn

nn < 0.

Since (n + 1) < (n + 2)

un - un+1 < 0 un < un+1

further n

lim un = n

lim 1 -

)1log(

1

n = 1 – 0 = 1 0.

Both the conditions of the Leibnitz test are not satisfied. So, we conclude that the series

oscillates between - and + .

Problems:

Test the convergence of the following series

1

1

1

2

2

1 1 11 ....

2 3 4

1 1 1 1....

log 2 log3 log 4 log5

1( )

1

1 0 1

1

1

1

n

n

n n

n

i

ii

niii

n

xiv for x

n n

vn

ABSOLUTELY & CONDITIONALLY CONVERGENT SERIES


1 1

1n

n n

n n

a u

is said to be absolutely convergent if the positive

series 1 2 3 4 ... na a a a a is convergent.


1 1

1n

n n

n n

a u

is said to be conditionally convergent if

(i) na is divergent

(ii) 1

1 1

1n

n n

n n

a u

is convergent

Theorem: An absolutely convergent series is convergent. The converse need not be true.

Proof: Let 1

1 1

1n

n n

n n

a u

be an absolutely convergent series then na is

convergent.

We know, 1 2 3 4 1 2 3 4... ...a a a a a a a a

By comparison test, 1

n

n

a

is convergent.

Q. Show that each of the following series also converges absolutely

(i) an2; (ii)

2

2

1 n

n

a

a

; (iii)

n

n

a

a

1

Solution: (i) Since an converges, we have an 0 as n . Hence for some positive

integer N, |an| < 1 for all n N. This gives an2 |an| for all n N. As |an| is convergent it

follows an2 converges.

(as an2 is a positive termed series, convergence and absolute convergence are identical).

(ii) As 1 + an2 1 for all n, we get

2

2

1 n

n

a

a

an

2

the convergence of an2 implies the convergence of

2

2

1 n

n

a

a

.

(iii) n

n

a

a

1 =

|1|

||

n

n

a

a

<

||1

||

n

n

a

a

.

As |an| converges, |an| 0 as n . Hence for some positive integer N, we have |an| < ½

for all n N.

This gives n

n

a

a

1 < 2|an| for all n N.

Now, by comparison test, n

n

a

a

1 converges.

That is, n

n

a

a

1converges absolutely.

Q. Test the convergence 3 3 3 3

1 1 1 11 2 1 2 3 1 2 3 4 ...

2 3 4 5

Solution: Here

1 1 1

3 2

1 2 ...1 1 1

1 2 1

n n n

n n

n na u

n n

then

2

1 2 2

1 10

2 1 2n n

n nu u

n n

i.e., 1 &lim 0n n n

nu u u

Thus by Lebinitz rule, na is convergent.

Also, 2

1

2 1n

na

n

. Take

1nv

n

Then 1

lim 02

n

nn

a

v

Since is nv divergent, therefore na is also divergent.

Thus the given series is conditionally convergent.

POWER SERIES

A series of the form 2

0 1 2 ... ... ( )n

na a x a x a x i where the ia ’s are

independent of x, is called a power series in x. Such a series may converge for some or all

values of x.

INTERVAL OF CONVERGENCE

In the power series (i) we have n

n nu a x

Therefore, 1 1lim limn n

n nn n

u ax

u a

If 1lim n

nn

al

a

, then by ratio test, the series (i) converges when

1x

l and diverges for

other values.

Thus the power series (i) has an interval 1 1

xl l

within which it converges and diverges

for values of x outside the interval. Such interval is called the interval of convergence of the

power series.

Q. Find the interval of convergence of the series 2 3 4 5

....2 3 4 5

x x x xx .

Solution: Here 1

1n

n

n

xu

n

and

1

1 11

nn

n

xu

n

Therefore, 1lim lim1

n

n nn

u nx x

u n

By Ratio test the given series converges 1x for and diverges for 1x .

When x=1 the series reduces to 1 1 1

1 ...2 3 4

, which is an alternating series and is

convergent.

When x=-1 the series becomes1 1 1

1 ...2 3 4

, which is divergent (by comparison with

p-series when p=1)

Hence the interval of convergence is 1 1x .

Q. Show that the series

1

1)1( n

12 n

x n

is absolutely convergent for | x | < 1,

conditionally convergent for x = 1 and divergent for x = -1.

Solution. Here un = (-1)n-1

12 n

x n

Therefore un+1 = 32

)1( 1

n

x nn

nlim

n

N

u

u 1 = n

lim nn

nn

xn

nx1

1

)1(32

12)1(

= n

lim xn

n

32

12)1(

= n

lim xnn

nn

)/32(

)/12()1(

= | x |

Therefore by generalized D’ Alembert’s test the series is absolutely convergent if

| x | < 1, not convergent if | x | > 1 and the test fails if | x | = 1.

Now for |x | = 1, x can be +1 or – 1.

If x = 1 the given series becomes 3

1 -

5

1+

7

1 -

9

1 + …

Here un =12

1

n, un+1 =

32

1

n

But 2n + 1< 2n + 3 un > un+1

Also n

lim un = n

lim 32

1

n = 0

Therefore by Leibnitz test the series is convergence when x = 1.

But the absolute series 3

1 +

5

1 +

7

1 + … whose general term is un =

12

1

n and is of

order n

1 =

2/1

1

n and hence un is divergent

Since the alternating series is convergent and the absolute series is divergent when x = 1, the

series is conditionally convergent when x = 1.

If x = -1, the series becomes 3

1 -

5

1 -

7

1 - ….

= -

...

7

1

5

1

3

1 where the series of positive terms is divergent as shown already.

Therefore the given series is divergent when x = -1.

Thus we have established all the results.

Problems:

1. Test the conditional convergence of

1 1

1 2

1 1( )

1

n n

n n

ni ii

nn

2. Prove that 3 3 3

sin sin 2 sin3....

1 2 3

x x x is absolutely convergent

3. For what values of x the following series are convergent

2 2

2 3 4

2 3 4

2 2 2

2 3 4

4 4 9 9

....2 3 4

....2 3 4

....1.2 2.3 3.4 4.5

3 3 3 .... 3 ...n n

x x xi x

x x xii x

x x x xiii

iv x x x x

4. Test the nature of convergence

11

n

n n

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