Series Module 4 -September 2010

29/12/20105. Series I 1

Infinite Series

Module 4

29/12/20105. Series I 2

Learning Objectives

At the end of the module, students able to

• determine whether a series converges or

diverges using suitable technique for test

of convergence.

29/12/20105. Series I 3

Techniques to determine whether

series converge or diverge.na

29/12/20105. Series I 4

Theorem

If the series is convergent, then 1n

na .0lim nn

a

Not always

true

29/12/20105. Series I 5


na .0lim nn

a

Always true

Example:

12

3

n n

03

lim2nn

then1

2

3

n nconverges,

29/12/20105. Series I 6


na .0lim nn

a

Not always true

Example: The Harmonic Series

1

3

n n

03

limnn

however,1

3

n ndiverges.

29/12/20105. Series I 7

The Test for Divergence

If does not exist or if then the

series is divergent.

nn

alim ,0lim nn

a

1n

na

29/12/20105. Series I 8

Example 1

Show that the series diverges. 1

2

2

45n n

n

29/12/20105. Series I 9

Example 1 - solution

05

1

5

1lim

45lim

2

42

2

nnn n

n

By the Test for Divergence, the series

diverges.

Reminder:What is the difference between sequenceand series?

29/12/20105. Series I 10

Theorem

If and are convergent series, then so are

the series

na nb

),(),(, nnnnn babaca and and

1 11

1 11

1 1

)()iii(

)()ii(

)i(

n n

nn

n

nn

n n

nn

n

nn

n n

nn

baba

baba

acca

29/12/20105. Series I 11

From the Test for Divergence

• If the series may converge or diverge.

• If the series must diverge.

0lim nn

a

0lim nn

a

29/12/20105. Series I 12

Example 2

1

2n

n

12

2

12

3

n n

nn

12

2

45n n

n

Determine whether the following series converge or

diverge?

(i)

(ii)

(iii)

29/12/20105. Series I 13


11

1

)2(2nn

n n

12

2

12

3

n n

nn

(i)

(ii)

012)2(lim 01n

n, the series diverges

02

1

2lim

12

3lim

2

2

2

2

n

n

n

nn

nn

, the series

diverges.

29/12/20105. Series I 14


(iii)

05

1

5

1lim

5lim

45lim

22

2

nnn n

n

n

n

Therefore, the series diverges.

12

2

45n n

n

12

2

45n n

n

29/12/20105. Series I 15

Practice

(i)

1

)1

1(n

n

n

By using the divergence test, show that the

series diverges.

(ii)

1 3k k

k

1kke

k(iii)

29/12/20105. Series I 16

Remark

If , then the series may converges

or diverges.

Therefore, need to use other test techniques

to confirm.

0lim nn

a

29/12/20105. Series I 17

Exercise

Exercise 12.2, Page 756No. 21, 25, 29, 32

29/12/20105. Series I 18

The Comparison Test

A series of positive terms is convergent if its terms

are less than the corresponding terms of a positive

series which is known to be convergent.

The BIG…

convergeThe

SMALL

subset

MUST

converge

29/12/20105. Series I 19

The Comparison Test

Similarly, the series is divergent if its terms are

greater than the corresponding terms of a series

which is known to be divergent.The SMALL

diverge.The

BIG

???

29/12/20105. Series I 20

Example 4

Determine whether the series

12 342

5

n nn

converges or diverges.

29/12/20105. Series I 21


22 2

5

342

5

nnn(based on the

denominator)

Since the series is convergent.1

21

2

1

2

5

2

5

nn nn

By Comparison Test

12 342

5

n nnconvergent.

WHY?

29/12/20105. Series I 22

Example 5

1

ln

n n

nDetermine whether converges

or diverges.

29/12/20105. Series I 23


3,1ln nn

nn

n 1ln

The series is divergent. (Harmonic series) 1

1

n n

By Comparison Test

1

ln

n n

ndiverges.

29/12/20105. Series I 24

Practice

By the Comparison Test, determine whether

the series converges or diverges.

1 21

14

4)ii(

4

2)i(

n

n

k

n

29/12/20105. Series I 25

Remark 1

There are two steps required for the comparisontest to determine whether a series withpositive term converges or diverges:

ka

• Guess whether the series converges or diverges .

• Find a series that proves the guess is correct.

ka

ku

29/12/20105. Series I 26

Remark 1 - continue

Find a series that proves the guess is correct.

1. If the guess that is divergence, must finda divergence series where .

2. If the guess that is convergence,must finda convergent series whose terms

ku

ka

kukk ua

ka

.kk auku

29/12/20105. Series I 27

Remark 2

The Comparison Test only applies to series

with nonnegative terms.

Try this test as last resort; other tests are

often easier to apply.

29/12/20105. Series I 28

The Limit Comparison Test

Given the positive terms series

1n

naand

.1n

nb

If cb

a

n

n

nlim

where c is a finite number and c > 0, then either

both series converge or both diverge.

29/12/20105. Series I 29

Example 6

By the Limit Comparison Test, determine

whether the series, converges or diverges.

1 12

1

nn

29/12/20105. Series I 30


1 12

1

nn

Step 1: Find a new series with from the given series.

1 2

1

nn

an

bn

29/12/20105. Series I 31


1 12

1

nn

Step 2: Compute

1 2

1

nn

an bn

n

n

n b

alim

0112

2lim

2

112

1

limlimn

n

n

n

n

nn

n

n b

a

29/12/20105. Series I 32


1 12

1

nn

1 2

1

nn

Since the series is a convergent geometric

series, by the Limit Comparison Test, the series

1 2

1

nn

1 12

1

nn

converges.

29/12/20105. Series I 33

Example 7

15

2

5

32

k k

kk

By the Limit Comparison Test, determine whether

the series converges or diverges.

29/12/20105. Series I 34

Example 7- solution

15

2

5

32

k k

kk

The highest power of the numerator is 2k2 and

the highest power of the denominator is k5/2 .

Thus,

21

22and

5

32

5

2

5

2

kk

kb

k

kka kk

29/12/20105. Series I 35

Example 7- solution

21

22and

5

32

5

2

5

2

kk

kb

k

kka kk

Then

)2

.(5

32lim

25

32

lim21

21

5

25

2

k

k

kk

k

k

kk

kk

1

15

2

32

lim52

32lim

5

5

23

25

k

k

k

kk

kk

29/12/20105. Series I 36

Example 7- solution

21

22and

5

32

5

2

5

2

kk

kb

k

kka kk

Since it is divergent p-series,

therefore by the Limit Comparison Test, 1

5

2

5

32

k k

kk

diverges.

29/12/20105. Series I 37

Practice

By the Limit Comparison Test, determine whether

the following series converges or diverges.

1

1k7

2

13

5 (ii)

88

624 (i)

kk

kk

kk

29/12/20105. Series I 38

Remark

The Limit Comparison Test is easier to apply than

the Comparison Test, however it requires skill in

choosing the series for comparison.1n

nb

We must know whetherthis series converges ordiverges.

29/12/20105. Series I 39

Exercise

Exercise 12.4, Page 770 – 771

No. 1, 3-32 (odd numbers only)

29/12/20105. Series I 40

The Integral Test

Suppose f (x) is a continuous, positive, decreasing

function on and let ),1[ ).(nfan

Then the series is convergent if and only if

the improper integral is convergent.

That is

1n

na

1

)( dxxf

29/12/20105. Series I 41

The Integral Test

1n

na

1

)( dxxf(i) If is convergent, then is

convergent.

(ii) If is divergent, then is

divergent.

1

)( dxxf1n

na

29/12/20105. Series I 42

Example

Test the series

12 1

1

n n

for convergence or divergence.

29/12/20105. Series I 43

Example - solution

Let

12 1

1

n n

,1

1)(

2xxf

We know that this function is continuous,

positive, and decreasing on

So, can use the Integral Test.

).,1[

29/12/20105. Series I 44

Example - solution

t

tdx

xdx

x1

2

1

2 1

1lim

1

1

t

tx

1

1tanlim

4421tantanlim 11 t

t

29/12/20105. Series I 45

Example - solution

1

2 1

1dx

xThus, is convergent integral,

so by the Integral Test, the series

is convergent.

12 1

1

n n

29/12/20105. Series I 46

Exercises

Exercise 12.3, Page 765

No. 1-24 (odd only)

29/12/20105. Series I 47

An Alternating Series

An alternating series is a series whose terms

are alternately positive and negative.

...6

1

5

1

4

1

3

1

2

11

)1(

1

1

n

n

n

29/12/20105. Series I 48

The Alternating Series Test

(AST)

If the alternating series

.convergent is series the then

(ii)

all fori.e., (i)

satisfies

n0lim

)...(,

)0(...)1(

1231

4321

1

1

n

nn

n

n

n

n

b

nbbbbb

bbbbbb

29/12/20105. Series I 49

Example 1

...6

1

5

1

4

1

3

1

2

11

)1(

1

1

n

n

n

Determine whether the series converge or

diverge by AST.

29/12/20105. Series I 50


...6

1

5

1

4

1

3

1

2

11

)1(

1

1

n

n

n

01

limlim (ii)

1

1

1because (i)

n

1

nb

nnbb

nn

nn

Therefore, by the Alternating Series Test

(AST) the series is convergent.

29/12/20105. Series I 51

Example 2

Determine whether the series converges

or diverges.

1

1

)1(

3)1(

k

k

kk

k

29/12/20105. Series I 52


1

1

)1(

3)1(

k

k

kk

k

(i) To show bk+1< bk

)1(

3

)2)(1(

4

kk

k

kk

k

29/12/20105. Series I 53



)1(

3

)2)(1(

4

kk

k

kk

k

)2)(1(

)3)(2(

)2)(1(

)4(

kkk

kk

kkk

kk

29/12/20105. Series I 54



)2)(1(

)65(4 22

kkk

kkkk

0)2)(1(

)6(

)2)(1(

6

kkk

k

kkk

k

therefore, bk+1< bk

29/12/20105. Series I 55


(ii) To show 0)1(

3lim

kk

k

k

01

lim3

lim)1(

3lim

1

31

2

2

k

kk

kkk kk

k

kk

k

since condition (i) and (ii) are satisfied, by

AST the series converges.

29/12/20105. Series I 56

Practice

By the Alternating Series Test, determine

whether the following series converges

or diverges.

1

1

12

1

)1( (ii)

)1( (i)

k

kk

n

n

e

n

29/12/20105. Series I 57

Remark

The Alternating Series Test only applies to

the alternating series.

29/12/20105. Series I 58

Exercises


No. 1, 2-20 (odd numbers only), 33

29/12/20105. Series I 59

Type of Convergence

For an alternating series, there are two types

of convergence:

(i) Absolutely Convergence;

(ii) Conditionally Convergence.

29/12/20105. Series I 60

Absolutely Convergence

A series is called absolutely

convergent

if the series of absolute value

is convergent.

1

)1(n

n

n a

1

)1(n

n

n a

29/12/20105. Series I 61

Example 1

The series ...4

1

3

1

2

11

)1(222

12

1

n

n

n

is absolutely convergent because

...4

1

3

1

2

11

1)1(222

12

12

1

nn

n

nnis convergent.

Why?

29/12/20105. Series I 62

Conditionally Convergence

A series is called conditionally

convergent if the series is convergent

however the series diverges

(i.e., not absolutely convergent).

n

n a)1(

1

)1(n

n

n a

29/12/20105. Series I 63

Example 2

The series ...4

1

3

1

2

11

)1(

1

1

n

n

n

is convergent , however the series

...4

1

3

1

2

11

1)1(

1 1

1

n n

n

nnis divergent.

...4

1

3

1

2

11

)1(

1

1

n

n

n

conditionally

convergent.

29/12/20105. Series I 64

Theorem

If a series is called absolutely convergent,

then the series is convergent.

1

)1(n

n

n a

thenconvergent absolutely is )1( If1n

n

n a1

)1(n

n

n a

is convergent.

29/12/20105. Series I 65

The Ratio Test

,1lim 1 La

a

n

n

n1

)1(n

n

n a(i) If then the series

is absolutely converges (and therefore

converges).

1

)1(n

n

n aGiven an alternating series

29/12/20105. Series I 66

The Ratio Test

,1lim 1 La

a

n

n

n1

)1(n

n

n a(ii) If then the series

diverges.

Note: The series maybe

convergent or divergent.

1

)1(n

n


1

)1(n

n

n a

29/12/20105. Series I 67

The Ratio Test

,1lim 1

n

n

n a

a(iii) If then the Ratio Test

is inconclusive.

1

)1(n

n


29/12/20105. Series I 68

Example 1

Using the ratio test, determine whether the

given series converge or diverge.

1

1

1

12

1 (iii)

4

)!2( (ii)

!

1 (i)

k

kk

n

k

k

n

29/12/20105. Series I 69

Example 1- solution

101

1lim

)!1(

!limlimlim

!

1 (i)

!1

)!1(1

1

1

n

n

n

a

a

n

n

nn

n

nn

n

n

n

By the Ratio Test, the series converges.

29/12/20105. Series I 70

Example 1- solution

By the Ratio Test, the series diverges.

)12)(22(lim4

1

4

1.

)!2(

)!22(lim

4

]!2[4

)]!1(2[

limlim

4

)!2( (ii)

11

1

kkk

k

k

k

a

a

k

kk

k

k

kk

k

k

kk

29/12/20105. Series I 71

Example 1- solution

By the Ratio Test, the series may be

converge or diverges.

112

12lim

12

1

1)1(2

1

limlim

12

1 (iii)

1

1k

k

k

k

k

a

a

k

kkk

k

k

29/12/20105. Series I 72

Example 2

Determine whether the series below, is

absolutely convergent, conditionally

convergent or divergent.

1

3

3)1(

nn

n n

29/12/20105. Series I 73


1

3

3)1(

nn

n nn

n

n

na

3)1(

3

31

3

3

1

31

1 3.

3

)1(lim

3)1(

3

)1()1(

limlimn

n

n

n

a

a n

nn

n

n

n

n

nn

n

n

29/12/20105. Series I 74


3

1lim

3

1

)1(

3

1lim

3.

3

)1(lim

3

3

3

3

31

3

n

n

n

n

n

n

n

n

n

nn

By the Ratio Test, the series converge

absolutely ( the series is also convergent).

29/12/20105. Series I 75

Remark

May like to use the Ratio Test,when an

involves factorials or nth power.

29/12/20105. Series I 76

Exercises

Exercise 12.6, Page 781-782

No. 1, 3,7,9

29/12/20105. Series I 77

Exercises

Use the Ratio Test to determine whether the

series converges. If the test is inconclusive,

then say so.

12

1

13

1

15 4. )

2

1( 3.

! 2.

!

4 1.

nk

k

nk

k

n

nk

n

n

k

29/12/20105. Series I 78

The Root Test

.convergent is

series then the,1lim If (i)1

nn

nn

n aLa

divergent. is

series then the,limor ,1lim If (ii)1

nnn

nn

nn

n aaLa

ve.inconclusi,1lim If (iii)n

nna

29/12/20105. Series I 79

Power Series

It is a series of the form

1n

n

n xa(i)

1

)1(n

n

n

n xa(ii)

(iii)

1

)()1(n

n

n

n cbxa

29/12/20105. Series I 80

Power Series

You need to know how to find

radius of convergence and interval of

convergence.

29/12/20105. Series I 81

Example 1

Use the Root Test to determine whether the

following series converge or diverge.

1

2

))1(ln(

1 (ii)

12

54 (i)

nn

k

k

n

k

k

29/12/20105. Series I 82


1212

54lim

12

54lim

12

54 (i)

2

k

k

k

k

k

k

k

k

k

k

k

k

By the Root Test the series diverges.

29/12/20105. Series I 83


10)1ln(

1lim

))1(ln(

1lim

))1(ln(

1 (ii)

1

nn

n

nn

nn

nn

By the Root Test the series converges

(Absolutely convergence).

29/12/20105. Series I 84

Example 2

Test the convergence of the series

.23

32)1(

1

1

k

k

k

k

k

29/12/20105. Series I 85


.23

32)1(

1

1

k

k

k

k

k

3

2

23

32lim

23

32lim

k

k

k

k

k

k

k

k

By the Root Test, the series absolutely converges

(and therefore converges).

29/12/20105. Series I 86

Remark

May like to use the Root Test, when an ,

involves nth powers.

29/12/20105. Series I 87

Exercises


No. 1, 2 – 28 ( odd numbers only),

29,31.

29/12/20105. Series I 88

Exercises


series converges. If the test is inconclusive,

then say so.

1 1

11

)1( 4. 6

3.

200 2.

12

23 1.

k n

kk

k

n

n

k

k

ek

n

k

k

29/12/20105. Series I 89

Practice

Using the Ratio Test, determine whether

the given series converge or diverge.

1

1

! (ii)

!

3 (i)

k

k

k

k

k

k

k

29/12/20105. Series I 90

Practice


series converges or diverges.

1

1

1 (ii)

12

13 (i)

k

k-k

n

n

)-e(

n-

n

29/12/20105. Series I 91

Practice

Classify whether each series is conditionally

convergent, absolutely convergent or

divergent.

14 3

1

1

1)1( (ii)

159

13)1( (i)

k

k

n

n

k

n

n

29/12/20105. Series I 92

END of MODULE

Series Module 4 -September 2010

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Transcript of Series Module 4 -September 2010