lectures 1 4

Mathematics I

Chapter 10

Dr. Devendra Kumar

Department of Mathematics

Birla Institute of Technology & Science, Pilani

20152016

Devendra Kumar BITS, Pilani Mathematics I

CHAPTER 10

Infinite Series


In this Chapter, we are going to learn about Infinite

series and their convergence properties. We will be

discussing the following topics:

Sequences (Certain Theorems on Sequences)






Infinite Series






Infinite Series

Integral Test






Infinite Series

Integral Test

Comparison Tests


Ratio and Root Tests



Alternating Series



Alternating Series

Power Series



Alternating Series

Power Series

Taylor & Maclaurin Series


Section 10.1

Sequences [Self Study]


Sequence

A function whose domain is the set of natural

numbers ( f :NR) is called a sequence of realnumbers. We write f (n)= an, then the sequence isdenoted by {a1,a2, . . .} or by {an}

n=1 or simply by {an}.

We call an the nth term of the sequence or the value

of the sequence at n.


Examples

{n}= {1,2,3, . . . ,n, . . .}


Examples

{n}= {1,2,3, . . . ,n, . . .}{1n

}=

{1, 1

2, 13, . . . , 1

n, . . .

}


Examples

{n}= {1,2,3, . . . ,n, . . .}{1n

}=

{1, 1

2, 13, . . . , 1

n, . . .

}{(1)n+1

}= {1,1,1,1, . . . , (1)n+1, . . .}


Examples

{n}= {1,2,3, . . . ,n, . . .}{1n

}=

{1, 1

2, 13, . . . , 1

n, . . .

}{(1)n+1

}= {1,1,1,1, . . . , (1)n+1, . . .}{

1 1n

}=

{0, 1

2, 23, . . . ,1 1

n, . . .

}


Convergent Sequence

A sequence {an} is said to converge to a number L, if

for every positive number , however small, we can

find a positive integer N (depending on ) such that

|anL| < , n>N.


If no such number L exists, the sequence is said to

diverge.



diverge.

If {an} converges to L, then we write limnan = L or

simply an L.



diverge.


simply an L.The number L is called the limit of the sequence.



diverge.


simply an L.The number L is called the limit of the sequence.

A sequence can not converges to more than one

limit i.e., Limit of a sequence is unique.


Graphical Representation


Examples

The sequence{1n

}converges to the number 0

(using Archimedean property).


Examples

The sequence{1n



The sequence {n} diverges.


Examples

The sequence{1n




The sequence{(1)nn




Examples

The sequence{1n




The sequence{(1)nn



The sequence{1+ 1

n




Examples

The sequence{1n




The sequence{(1)nn



The sequence{1+ 1

n



The sequence {(1)n} diverges.Devendra Kumar BITS, Pilani Mathematics I

Theorem (Theorem 1.)

Let {an} and {bn} be two convergent sequences. Then

1 limkan = k liman2 lim(anbn)= liman limbn3 lim(an.bn)= liman. limbn4 lim

(anbn

)= liman

limbn, if bn , 0 for all n and limbn , 0.


Theorem (Theorem 2. Sandwich Theorem or

Squeeze Theorem)

Let {an}, {bn}, and {cn} be sequences of real numbers.

If an bn cn for all nN and if liman = lim cn = L,then limbn = L.



Squeeze Theorem)



Q:63 Find lim n!nn

(If exists).



Squeeze Theorem)



Q:63 Find lim n!nn

(If exists).

Sol. 0< n!nn= n(n1)(n2)...3.2.1

n.n.n...n.n.n 1

n 0. So an

application of Sandwich theorem gives lim n!nn= 0.


Theorem (Theorem 3. Continuous Function

Theorem)

Let {an} be a sequence of real numbers. If an L andif f (x) is a function that is continuous at x= L anddefined at all an, then f (an) f (L).



Theorem)


Q:43 Find limsin(pi

2+ 1

n

)(If exists).



Theorem)


Q:43 Find limsin(pi

2+ 1

n

)(If exists).

Sol. Let an = pi2 + 1n . Then

liman = lim(pi

2+ 1n

)= limpi

2+ lim 1

n= pi2.

So using continuous function theorem

limsin(pi

2+ 1

n

)= sin pi

2= 1.



Suppose that f (x) is a function defined for all x n0and that {an} is a sequence of real numbers s.t.

an = f (n) for n n0, thenlimx f (x)= L limnan = L.


Q:50 Find lim lnnln2n

.


Q:50 Find lim lnnln2n

.

Sol.

limln x

ln2x= lim

1x

22x

, by LHospitals rule

= 1.

Hence lim lnnln2n

= 1.


Bounded Sequence

A sequence {an} is said to be bounded from above if

there exists a number M1 such that

an M1, n.The number M1 is called an upper bound for {an}.


Bounded Sequence

A sequence {an} is said to be bounded from above if

there exists a number M1 such that

an M1, n.The number M1 is called an upper bound for {an}.

Similarly, a sequence {an} is said to be bounded from

below if there exists a number M2 such that

an M2, n.The number M2 is called a lower bound for {an}.


Bounded Sequence

A sequence {an} is said to be bounded if there exists a

number M such that

|an| M, n.The number M is called a bound for {an}.

Examples

The sequence{

nn+1

}is bounded below by 1

2and

bounded above by 1.


Bounded Sequence

A sequence {an} is said to be bounded if there exists a

number M such that

|an| M, n.The number M is called a bound for {an}.

Examples

The sequence{

nn+1

}is bounded below by 1

2and

bounded above by 1.

The sequence{12n

}is bounded below by 0 and

bounded above by 12.


Nondecreasing/Nonincreasing/Monotonic

Sequence

A sequence {an} is called a nondecreasing sequence,

if an an+1 for all n. The sequence is nonincreasing,if an an+1 for all n. The sequence is monotonic if itis either nondecreasing or nonincreasing.



Sequence



Examples

The sequence{

nn+1

}is nondecreasing.



Sequence



Examples

The sequence{

nn+1

}is nondecreasing.

The sequence{12n

}is nonincreasing.


Theorem (Theorem 6. Monotonic Sequence

Theorem)

A bounded monotonic sequence is convergent.


Theorem (Theorem 6. Monotonic Sequence

Theorem)

A bounded monotonic sequence is convergent.

Example

The sequence{1 1

2n

}is monotonic (nondecreasing)

and bounded (as |an| 1), therefore it is convergent.


Homework

Using following two facts:

The converse of Theorem 6 is not true.

Every convergent sequence is bounded (Theorem).

Give a counter example of a convergent sequence

which is not monotonic.


Section 10.2

Infinite Series


Infinite series

An infinite series is the sum of the terms of a

sequence. Thus an infinite series consists of infinite

number of real numbers separated by + sign. Thusfor a sequence {an} the series will be written as

a1+a2+a3+ +an+ or asn=1

an or simply asan. The number an is called the n

th term of the

series.


Examples

1+ 12+ 1

3+ + 1

n+ (Harmonic Series)


Examples

1+ 12+ 1

3+ + 1


11+11+ + (1)n+1+


Examples

1+ 12+ 1

3+ + 1


11+11+ + (1)n+1+ 1+ 1

22+ 1

32+ + 1

n2+


Sequence of Partial Sums

Consider the seriesan. The sequence {Sn}, defined

by

Sn = a1+a2+a3+ +an, nis called the sequence of partial sums of the series.


Sequence of Partial Sums

Consider the seriesan. The sequence {Sn}, defined

by

Sn = a1+a2+a3+ +an, nis called the sequence of partial sums of the series.

The nth term Sn of the sequence is called nth partial

sum of the series.


Convergent Series

A seriesan is said to be convergent iff the sequence

{Sn} is convergent. If {Sn} converges to a real number

L, then the seriesan also converges to L and the

sum of the series is L.


Convergent Series

A seriesan is said to be convergent iff the sequence

{Sn} is convergent. If {Sn} converges to a real number

L, then the seriesan also converges to L and the

sum of the series is L.

A series which is not convergent is called divergent.


Telescoping Series 1

n(n+1)Here an = 1n(n+1) , the Sn can be explicitly evaluatedas follows:

an =1

n(n+1) =1

n 1n+1.

Thus

Sn =(1 1

2

)+

(1

2 13

)+ +

(1

n 1n+1

)= 1 1

n+1.

Thus limSn = 1 and hence 1

n(n+1) = 1.


Q:3 Find the formula for nth partial sum and use it

to find the series sum if the series converges:

1 12+ 14 18+ + (1)n1 1

2n1+


Q:3 Find the formula for nth partial sum and use it

to find the series sum if the series converges:

1 12+ 14 18+ + (1)n1 1

2n1+

Sol. Here Sn =1(1

(12

)n)1(12)

, therefore

limSn = 11(12)

= 23. Thus

1 12+ 14 18+ + (1)n1 1

2n1+ = 2

3.


Remark.

Adding or deleting a finite number of terms in a

series does not alter the behavior (convergence or

divergence) of the series. However the sum of the

series will change in the case of convergent series.


There are two well known series which will be used

quite often:



quite often:

1 The geometric series:arn1, a, 0.



quite often:

1 The geometric series:arn1, a, 0.

2 The p-series: 1

np, p R.


Theorem

The geometric seriesarn1 with a, 0, is

(i) convergent if |r| < 1 and arn1 = a1r .

(ii) divergent if |r| 1.


Theorem



(ii) divergent if |r| 1.Proof.Well prove the theorem by using the

definition of convergence. We have

Sn = a+ar+ar2+ . . .+arn1 =a(1 rn)1 r , r , 1.


Theorem



(ii) divergent if |r| 1.Proof.Well prove the theorem by using the

definition of convergence. We have

Sn = a+ar+ar2+ . . .+arn1 =a(1 rn)1 r , r , 1.

(i) First we consider the case when |r| < 1, in thiscase rn 0 as n. Therefore, Sn a1r as n,and hence the GS is convergent. Also, its sum in this

case is a1r .


(ii) Now we consider the case when |r| 1. Again wesplit this case as |r| > 1 and |r| = 1.


(ii) Now we consider the case when |r| 1. Again wesplit this case as |r| > 1 and |r| = 1.Let we first consider |r| > 1: Then r > 1 or r

(ii) Now we consider the case when |r| 1. Again wesplit this case as |r| > 1 and |r| = 1.Let we first consider |r| > 1: Then r > 1 or r 1: We have rn as n and hence Snalso tends to or (depending on the sign ofa).


(ii) Now we consider the case when |r| 1. Again wesplit this case as |r| > 1 and |r| = 1.Let we first consider |r| > 1: Then r > 1 or r 1: We have rn as n and hence Snalso tends to or (depending on the sign ofa).

For r

Now consider |r| = 1: Then r = 1 or r =1.


Now consider |r| = 1: Then r = 1 or r =1.If r = 1, then GS becomes a+a+ . . ., so thatSn = na; this tends to or as n(depending on the sign of a); and hence GS is

divergent in this case also.


Now consider |r| = 1: Then r = 1 or r =1.If r = 1, then GS becomes a+a+ . . ., so thatSn = na; this tends to or as n(depending on the sign of a); and hence GS is

divergent in this case also.

If r =1, then GS becomes aa+aa+ . . ., so that

Sn ={0, if n is even

a, if n is odd.

Since a is not zero, Sn does not tend to a unique

limiting value and hence is divergent in this case

also. Devendra Kumar BITS, Pilani Mathematics I

Q:50n=0

(p2)n.


Q:50n=0

(p2)n.

Sol. Here r =p2> 1, therefore

n=0

(p2)n is

divergent.


Q:50n=0

(p2)n.


n=0

(p2)n is

divergent.

Q:57 2

10n.


Q:50n=0

(p2)n.


n=0

(p2)n is

divergent.

Q:57 2

10n.

Sol. Here r = 110< 1, therefore 2

10nis convergent

and 210n

=210

1 110

= 29.


Q:67n=0

(epi

)n.


Q:67n=0

(epi

)n.

Sol. Here r = epi< 1, therefore n=0 ( epi)n is convergent

and n=0

( epi

)n= 11 e

pi

= pipi e .


Q:67n=0

(epi

)n.


and n=0

( epi

)n= 11 e

pi

= pipi e .

Q:68n=0

enpi

pine .


Q:67n=0

(epi

)n.


and n=0

( epi

)n= 11 e

pi

= pipi e .

Q:68n=0

enpi

pine .

Sol. Here r = epipie > 1 (How ?), therefore

n=0

enpi

pine is

divergent.


Theorem (Theorem 7. The necessary condition

for a series to be convergent)

Ifan converges, then liman = 0.


Theorem (Theorem 7. The necessary condition

for a series to be convergent)

Ifan converges, then liman = 0.

Proof.We have

a1 = S1an = SnSn1 for n= 2,3,4, . . .

If the given series converges then Sn L. Therefore,liman = lim(SnSn1) LL= 0.


Remark.

The converse of above theorem is not true

i.e., liman = 0 does not necessarily imply that theseries

an converges.


Remark.

The converse of above theorem is not true

i.e., liman = 0 does not necessarily imply that theseries

an converges.

Example

For the Harmonic series 1

n; liman = 0 but the series

is divergent (well see later). Thus we have following

important test for the divergence.


The nth term test for divergence

If liman fails to exist or is different from zero, thenan is divergent.




Q:34 or 53n=0

cosnpi.




Q:34 or 53n=0

cosnpi.

Sol. Here liman = limcosnpi= lim(1)n, which doesnot exist, therefore

n=0

cosnpi is divergent.


Q:60(

1 1n

)n.


Q:60(

1 1n

)n.

Sol. liman = lim(1 1

n

)n = lim(1+ 1n

)n = e1 , 0,therefore

(1 1

n

)nis divergent.


Q:60(

1 1n

)n.


n

)n = lim(1+ 1n


(1 1

n

)nis divergent.

Q:62 nn

n!.


Q:60(

1 1n

)n.


n

)n = lim(1+ 1n


(1 1

n

)nis divergent.

Q:62 nn

n!.

Sol. liman = lim nn

n!= (How?), therefore nn

n!is

divergent.


Q:66ln

(n

2n+1).


Q:66ln

(n

2n+1).

Sol. lim n2n+1 = lim 12+ 1n =

12, therefore

liman = limln(

n2n+1

)= ln 1

2, 0, and so

ln

(n

2n+1)is

divergent.



Ifan = A and

bn =B are two convergent series,

then



Ifan = A and

bn =B are two convergent series,

then(anbn)=

an

bn = AB.

kan = kan = kA (for any real number k).


Section 10.3

The Integral Test


For next few classes, series will be assumed to be

series of nonnegative terms i.e., an 0 for all n. Thismeans that Sn Sn+1 for all n. That is the sequence{Sn} is nondecreasing.


For next few classes, series will be assumed to be

series of nonnegative terms i.e., an 0 for all n. Thismeans that Sn Sn+1 for all n. That is the sequence{Sn} is nondecreasing.

Corollary (Corollary of Theorem 6)

A seriesan of nonnegative terms converges iff Sn is

bounded from above.


Theorem (Theorem 9. The Integral Test)

Letan be a series of positive terms. Let f (x) be a

positive, continuous and decreasing function for all

xN for some N; and let f (n)= an for all n. Thenthe series

n=N

an and the integralN f (x)dx both

converge or both diverge.


Theorem (Theorem 9. The Integral Test)

Letan be a series of positive terms. Let f (x) be a

positive, continuous and decreasing function for all

xN for some N; and let f (n)= an for all n. Thenthe series

n=N

an and the integralN f (x)dx both

converge or both diverge.

Remark.

The sum of the series and the value of integral need

not necessarily be equal in the convergent case.


Let we apply this test to p-series:

An Important Result

The p-series 1

npis convergent for p> 1, and

divergent for p 1.



An Important Result

The p-series 1


divergent for p 1.Proof. For p 0, the series is trivially divergent (asliman , 0).



An Important Result

The p-series 1


divergent for p 1.Proof. For p 0, the series is trivially divergent (asliman , 0).

So we consider p> 0. Note that for p> 0 the functionf (x)= 1

xpis positive, continuous and decreasing when

x 1. So we can apply integral test:


If p> 1, then1 f (x)dx =

1

1xpdx= limb

[xp+1p+1

]b1= 1

p1.Thus, the series converges in this case.



1

1xpdx= limb

[xp+1p+1

]b1= 1


If 0< p 1. We split the case



1

1xpdx= limb

[xp+1p+1

]b1= 1


If 0< p 1. We split the caseIf p= 1, we get

1

1xdx= limb[ln x]b1. So the

series is divergent.



1

1xpdx= limb

[xp+1p+1

]b1= 1



1



If 0< p< 1, then1 f (x)dx=

1

1xpdx= limb

[xp+1p+1

]b1. Hence the

p-series diverges in this case also.



1

1xpdx= limb

[xp+1p+1

]b1= 1



1



If 0< p< 1, then1 f (x)dx=

1

1xpdx= limb

[xp+1p+1

]b1. Hence the

p-series diverges in this case also.

An Important Result

By the above test the Harmonic series 1

ndiverges.


Q:18 Decide whether the series converges/diverges:

lnnn2

.


Q:18 Decide whether the series converges/diverges:

lnnn2

.

Sol. The function f (x)= ln xx2

is positive, continuous

and decreasing for all x 2 (note that these threeconditions holds true for all x> pe). Now, we have


2

ln x

x2dx=

ln2

tet dt (on substituting ln x= t)

= limb

[(t+1)et]bln2

= 1+ ln22

.

Hence by integral test the series lnn

n2is convergent.


Q:32 1

n(1+ln2n).


Q:32 1

n(1+ln2n).

Sol. The function f (x)= 1x(1+ln2 x) is positive,

continuous and decreasing for all x 1.1

1

x(1+ ln2 x)dx=

0

dt

1+ t2 (on substituting lnx= t)

= limb

[tan1 t]b0

= pi2.

Hence by integral test the series is convergent.


HomeworkWhich of the series converge and which

diverge? 1n.



diverge? 1n. 1

n2.



diverge? 1n. 1

n2.

Ans. Divergent, Convergent.



diverge? 1n. 1

n2.


Homework For what p> 0, does the series2

1n(lnn)p

converge?



diverge? 1n. 1

n2.


Homework For what p> 0, does the series2

1n(lnn)p

converge?

Ans. p> 1.


Section 10.4

Comparison Tests


Now, well study two comparison tests:



The Direct Comparison Test (DCT)



The Direct Comparison Test (DCT)

The Limit Comparison Test (LCT)


Theorem (Theorem 10. The Direct Comparison

Test)

Letan be a series of nonnegative terms. Then

(a) If an cn for all nN, wherecn is a

known convergent series; thenan

converges.


Theorem (Theorem 10. The Direct Comparison

Test)

Letan be a series of nonnegative terms. Then

(a) If an cn for all nN, wherecn is a

known convergent series; thenan

converges.

(b) If an dn for all nN, wheredn is a

known divergent series of nonnegative terms,

thenan diverges.


Example

As an application, well prove that the exponential

series 1

n!is convergent.


Example

As an application, well prove that the exponential

series 1

n!is convergent.

Sol. Here, an = 1n! . We know that 2n1 n! for all n.Therefore, we have,

an =1

n! 12n1

= cn (say) for all n.

Moreover, 1

2n1 , being a geometric series with r =12,

is convergent. Therefore, by DCT the exponential

series is convergent.


Q:18 3

n+pn


Q:18 3

n+pnSol. Here an = 3n+pn

3/2n

for all n. Since the series 1ndiverges, therefore by DCT

3n+pn diverges.


Q:18 3


3/2n


3n+pn diverges.

Q:26 1p

n3+2


Q:18 3


3/2n


3n+pn diverges.

Q:26 1p

n3+2Sol. Here an = 1p

n3+2 1

n3/2for all n. Since the series 1

n3/2converges, therefore by DCT

1pn3+2

converges.


Q:44 (n1)!

(n+2)!.


Q:44 (n1)!

(n+2)!.Sol. Here an = (n1)!(n+2)! = 1(n+2)(n+1)n < 1n3 for all n. Sincethe series

1n3

converges, therefore by DCT (n1)!

(n+2)!converges.


Theorem (11. Limit Comparison Test)

Suppose an and bn are positive for all nN.(a) If lim

anbn= c (finite and non zero), then both

an andbn converge or diverge together.






(b) If limanbn= 0, and bn converges, then an

also converges.






(b) If limanbn= 0, and bn converges, then an

also converges.

(c) If limanbn=, and bn diverges, then an

also diverges.


Q:24

n=35n33n

n2(n2)(n2+5).


Q:24

n=35n33n

n2(n2)(n2+5).

Sol. Here an = 5n33n

n2(n2)(n2+5). Consider bn =n3

n5= 1

n2.

Note thatbn is convergent. Now

liman

bn= lim

5n33nn2(n2)(n2+5)

1n2

= 5.

Hencean is convergent, by LCT (a).


Q:36 n+2n

n22n


Q:36 n+2n

n22n

Sol. Here an = n+2n

n22n. Since 2n > n for all n so consider

bn = 2n

n22n= 1

n2. Note that

bn is convergent. Now

liman

bn= lim

n+2nn22n

1n2

= lim n+2n

2n= 1.

Hencean is convergent, by LCT (a).


Q:28 (lnn)2

n3.


Q:28 (lnn)2

n3.

Sol. Here an = (lnn)2

n3. Consider bn = n

c

n3= 1

n3c , where c

is very small (close to zero). Note thatbn is

convergent. Now

liman

bn= lim

(lnn)2

n3

1n3c

= lim (lnn)2

nc= 0.

Hencean is convergent, by LCT (b).


Q:46tan 1

n


Q:46tan 1

n

Sol. Note the series of tan x

tanx= x+ 13x3+ 2

15x5+

Here an = tan 1n . Consider bn = 1n . Note thatbn is

divergent. Now

liman

bn= lim

tan 1n

1n

= 1.

Hencean is divergent, by LCT (a).


Section 10.5



Theorem (Theorem 12. Ratio Test)

Letan be a series of positive terms. Suppose

liman+1an

= r. Then




liman+1an

= r. Then(a) If r < 1, the series is convergent;




liman+1an

= r. Then(a) If r < 1, the series is convergent;(b) If r > 1, the series is divergent;




liman+1an

= r. Then(a) If r < 1, the series is convergent;(b) If r > 1, the series is divergent;(c) If r = 1, the test is inconclusive.


Q:32 n lnn

2n.


Q:32 n lnn

2n.

Sol. Here an = n lnn2n . So


Q:32 n lnn

2n.


an+1an

=(n+1)ln(n+1)

2n+1n lnn2n

= (n+1)ln(n+1)2n lnn

.

Thus

liman+1an

= 12< 1 (on applying LHospitals rule).


Q:32 n lnn

2n.


an+1an

=(n+1)ln(n+1)

2n+1n lnn2n

= (n+1)ln(n+1)2n lnn

.

Thus

liman+1an

= 12< 1 (on applying LHospitals rule).

Therefore by ratio test, the series converges.


Q:38 n!

nn.


Q:38 n!

nn.

Sol. Here an = n!nn . So


Q:38 n!

nn.


an+1an

=(n+1)!

(n+1)n+1n!nn

=( nn+1

)n=

(1 1

n+1

)n

liman+1an

= lim(1 1

n+1

)n= e1 < 1.


Q:38 n!

nn.


an+1an

=(n+1)!

(n+1)n+1n!nn

=( nn+1

)n=

(1 1

n+1

)n

liman+1an

= lim(1 1

n+1

)n= e1 < 1.

Therefore by ratio test, the series converges.


Q:42 3n

n32n.


Q:42 3n

n32n.

Sol. Here an = 3n

n32n. So


Q:42 3n

n32n.

Sol. Here an = 3n

n32n. So

an+1an

=3n+1

(n+1)32n+13n

n32n

= 32.

n3

(n+1)3

liman+1an

= lim 32.

n3

(n+1)3 =3

2> 1.


Q:42 3n

n32n.

Sol. Here an = 3n

n32n. So

an+1an

=3n+1

(n+1)32n+13n

n32n

= 32.

n3

(n+1)3

liman+1an

= lim 32.

n3

(n+1)3 =3

2> 1.

Therefore by ratio test, the series diverges.


Theorem (Theorem 13. Root Test or nth Root

Test)

Letan be a series of non-negative terms, and

suppose that lim(an)1n = r. Then

(a) If r < 1, the series is convergent;



Test)



(a) If r < 1, the series is convergent;(b) If r > 1, the series is divergent;



Test)



(a) If r < 1, the series is convergent;(b) If r > 1, the series is divergent;(c) If r = 1, the test is inconclusive.


Q:18n2en.


Q:18n2en.

Sol. Here

an = n2en

(an)1/n = n2/ne1

lim(an)1/n = limn2/ne1 = e1 < 1.


Q:18n2en.

Sol. Here

an = n2en

(an)1/n = n2/ne1

lim(an)1/n = limn2/ne1 = e1 < 1.Therefore the series converges by root test.


Q:40

n=2n

(lnn)(n/2).


Q:40

n=2n

(lnn)(n/2).

Sol. Here

an =n

(lnn)(n/2)

(an)1/n = n

1/n

(lnn)1/2

lim(an)1/n = lim n

1/n

(lnn)1/2= 0< 1.


Q:40

n=2n

(lnn)(n/2).

Sol. Here

an =n

(lnn)(n/2)

(an)1/n = n

1/n

(lnn)1/2

lim(an)1/n = lim n

1/n

(lnn)1/2= 0< 1.

Therefore the series converges by root test.


Section 10.6Alternating Series, Absolute and

Conditional Convergence


Alternating Series

A series whose terms are alternately positive and

negative is called an alternating series.


Alternating Series

A series whose terms are alternately positive and

negative is called an alternating series.

An alternating series is one of the form(1)n+1un

or(1)nun, where un > 0 for all n.


Examples(1)n+1


Examples(1)n+1(1)n+1 1

n(Alternating harmonic series)




(1)n ln(1+ 1

n

)




(1)n ln(1+ 1

n

)(1)n n

n+1


Theorem (Theorem 14. Leibnizs Theorem for

Alternating Series)

The alternating series(1)n+1un converges if

(i) un un+1 for all nN for some N; and



Alternating Series)


(i) un un+1 for all nN for some N; and(ii) un 0.



Alternating Series)


(i) un un+1 for all nN for some N; and(ii) un 0.

Proof. It is enough to show that the sequence Sn is

convergent (definition of convergence of a series). In

order to show that Sn converges, we use the

following result:


Ex. 131 on p. 543: For a sequence {Sn}, if {S2n} and

{S2n+1} converge to the same number L, then Sn L.



{S2n+1} converge to the same number L, then Sn L.First consider the sequence {S2n}. Well see that S2nconverges (by showing that S2n is non-decreasing

and bounded from above).



{S2n+1} converge to the same number L, then Sn L.First consider the sequence {S2n}. Well see that S2nconverges (by showing that S2n is non-decreasing

and bounded from above).

S2n is non-decreasing We have

S2n+2 = S2n+ (u2n+1u2n+2).Since u2n+1u2n+2 0, so S2n+2 S2n. Thus thesequence {S2n} is non-decreasing.


S2n is bounded from above Arrange S2n as

S2n = u1 (u2u3) (u2n2u2n1)u2n u1.Thus the sequence {S2n} is bounded from above.


S2n is bounded from above Arrange S2n as

S2n = u1 (u2u3) (u2n2u2n1)u2n u1.Thus the sequence {S2n} is bounded from above.

Therefore it is convergent and so has a limit, say L

i.e., limS2n L.


Now consider the sequence {S2n+1}. We have



S2n+1 = S2n+u2n+1 limS2n+1 = limS2n+ limu2n+1.




Therefore, using condition (ii):





limS2n+1 = L+0= L.





limS2n+1 = L+0= L.Thus Sn L and hence

(1)n+1un converges to

L.


Q:8(1)n 10n

(n+1)!.


Q:8(1)n 10n

(n+1)!.Sol. (i) Here un = 10

n

(n+1)!. Consider

unun+1 =10n

(n+1)!10n+1

(n+2)! =(n8)10n(n+2)! .

Thus unun+1 0 for all n 8 i.e., un un+1 for alln 8.


Q:8(1)n 10n

(n+1)!.Sol. (i) Here un = 10

n

(n+1)!. Consider

unun+1 =10n

(n+1)!10n+1

(n+2)! =(n8)10n(n+2)! .

Thus unun+1 0 for all n 8 i.e., un un+1 for alln 8.(ii) Also, we have limun = lim 10

n

(n+1)! = 0.


Q:8(1)n 10n

(n+1)!.Sol. (i) Here un = 10

n

(n+1)!. Consider

unun+1 =10n

(n+1)!10n+1

(n+2)! =(n8)10n(n+2)! .

Thus unun+1 0 for all n 8 i.e., un un+1 for alln 8.(ii) Also, we have limun = lim 10

n

(n+1)! = 0.Hence by Laibnizs test the series is convergent.


Q:12(1)n ln

(1+ 1

n

).


Q:12(1)n ln

(1+ 1

n

).

Sol. (i) Here un = ln(1+ 1

n

). Consider

f (x)= ln(1+ 1

x

) f (x)= 1

x(x+1).

Since f (x)< 0 for all x 1. So f (x) is decreasing forall x 1 and hence un un+1, n 1.


Q:12(1)n ln

(1+ 1

n

).


n

). Consider

f (x)= ln(1+ 1

x

) f (x)= 1

x(x+1).

Since f (x)< 0 for all x 1. So f (x) is decreasing forall x 1 and hence un un+1, n 1.(ii) Also, we have limun = limln

(1+ 1

n

)= 0 (how?).


Q:12(1)n ln

(1+ 1

n

).


n

). Consider

f (x)= ln(1+ 1

x

) f (x)= 1

x(x+1).

Since f (x)< 0 for all x 1. So f (x) is decreasing forall x 1 and hence un un+1, n 1.(ii) Also, we have limun = limln

(1+ 1

n

)= 0 (how?).

Hence by Laibnizs test the series is convergent.


Corollary

For alternating series, if limun does not tend to zero,

then the series diverges.


Corollary



Q:6(1)n+1n2+5

n2+4.


Corollary



Q:6(1)n+1n2+5

n2+4.

Sol. Here un = n2+5

n2+4. Since limun = 1, therefore byabove corollary series is divergent.


Absolute and Conditional Convergence

A seriesan is said to be absolutely convergent if

the series |an| is convergent.


Absolute and Conditional Convergence

A seriesan is said to be absolutely convergent if

the series |an| is convergent.

If the seriesan is convergent but not absolutely

convergent, then it is said to be conditionally

convergent.


Theorem (Theorem 16. The Absolute

Convergence Test)

If |an| converges, then an also converges. That is

absolute convergence implies convergence.


Theorem (Theorem 16. The Absolute

Convergence Test)

If |an| converges, then an also converges. That is

absolute convergence implies convergence.

Proof.We have

|an| an |an| , n.So

0 |an|+an 2 |an| .Now the series

2 |an| converges as the series

|an|converges.


Therefore by Direct Comparison Test, the

non-negative terms series(|an|+an) converges.


Therefore by Direct Comparison Test, the

non-negative terms series(|an|+an) converges.

Now we can writean =

(|an|+an|an|)=

(|an|+an)

|an| .

Thereforean, being a difference of two convergent

series converges.


Remark.

The converse of above theorem is not true.


Remark.

The converse of above theorem is not true.

Example

The alternating harmonic seriesan =

(1)n+1n

converges but |an| = 1n diverges.


Steps to check a series for absolute and conditional

convergence

First check the behavior of the series |an|. If |an| is convergent then an is absolutely

convergent.



convergence


convergent.

If |an| is divergent then check the behavior ofan. If

an is convergent then

an is

conditionally convergent.



convergence


convergent.

If |an| is divergent then check the behavior ofan. If

an is convergent then

an is

conditionally convergent.

Otherwisean is divergent.


Q:30(1)n lnn

nlnn.


Q:30(1)n lnn

nlnn.Sol.

First we check the behavior of |an| = lnnnlnn . We

havelnn

n lnn lnn

n 1n, n 3.

Also 1

nis divergent so

lnnnlnn is also divergent

(by DCT).


Now we check the behavior ofan =

(1)n lnn

nlnnby using Leibnizs test.



(1)n lnn


(i) Consider f (x)= ln xxlnx f (x)= 1lnx(xlnx)2 . Thus

f (x)< 0 for all x> e and so un un+1 for all n 3.



(1)n lnn



f (x)< 0 for all x> e and so un un+1 for all n 3.(ii) Also we have lim lnn

nlnn = 0 (Using LHospitalsrule).



(1)n lnn





Therefore by Leibnizs test(1)n lnn

nlnn isconvergent.



(1)n lnn





Therefore by Leibnizs test(1)n lnn

nlnn isconvergent.

Hence(1)n lnn

nlnn is conditionally convergent.


Q:38(1)n+1 (n!)2

(2n)!.


Q:38(1)n+1 (n!)2

(2n)!.

Sol. First we check the behavior of |an| = (n!)2(2n)!.

We have

|an+1||an|

=((n+1)!)2(2n+2)!(n!)2

(2n)!

= (n+1)2

(2n+1)(2n+2),

and so lim|an+1||an| =

14< 1.


Q:38(1)n+1 (n!)2

(2n)!.

Sol. First we check the behavior of |an| = (n!)2(2n)!.

We have

|an+1||an|

=((n+1)!)2(2n+2)!(n!)2

(2n)!

= (n+1)2

(2n+1)(2n+2),

and so lim|an+1||an| =

14< 1.

Therefore by ratio test |an| converges and hence

an is absolutely convergent.


Q:44(1)n 1p

n+pn+1.


Q:44(1)n 1p

n+pn+1.

Sol.

First we check the behavior of |an| = 1p

n+pn+1.

Consider bn = 1pn , then we have

liman

bn= lim

1pn+

pn+1

1pn

= 12.

Therefore by LCT (a) |an| is divergent.



(1)n 1p

n+pn+1 by using Leibnizs test.



(1)n 1p


(i) Since 1pn+

pn+1 >

1pn+1+

pn+2 for all n, so

un > un+1 for all n.



(1)n 1p


(i) Since 1pn+

pn+1 >

1pn+1+

pn+2 for all n, so

un > un+1 for all n.(ii) Also we have limun = lim 1p

n+pn+1 = 0.



(1)n 1p


(i) Since 1pn+

pn+1 >

1pn+1+

pn+2 for all n, so


n+pn+1 = 0.

Therefore by Leibnizs test(1)n 1p

n+pn+1 is

convergent.



(1)n 1p


(i) Since 1pn+

pn+1 >

1pn+1+

pn+2 for all n, so


n+pn+1 = 0.

Therefore by Leibnizs test(1)n 1p

n+pn+1 is

convergent.

Hence(1)n 1p

n+pn+1 is conditionally convergent.


Summary for Behavior of a Series

If liman9 0, the series diverges.




See if it is known series (like geometric series,

p-series) etc.





p-series) etc.

For nonnegative terms series try integral test,

comparison test, ratio test, root test etc.





p-series) etc.



For series with some negative terms (not

necessarily alternating series) see whether |an|

converges. If yes, so doesan.





p-series) etc.



For series with some negative terms (not

necessarily alternating series) see whether |an|

converges. If yes, so doesan.

For alternating series apply Leibnizs test.


Section 10.7

Power Series


A power series about x= a is a series of the formn=0

an(xa)n in which the center a andcoefficients an are constants.




A power series about x= 0 is a series which lookslike

n=0

anxn.




A power series about x= 0 is a series which lookslike

n=0

anxn.

In the expanded form, it looks like

a0+a1x+a2x2+ +anxn+ .Here x is a real variable.


Examplesn=0

xn (geometric series).


Examplesn=0


n=0

xn

n!(exponential series).


Examplesn=0


n=0

xn

n!(exponential series).

(1)n+1 xn

n(logarithmic series).


Remark.

A power series always converges at the center

i.e.,n=0

an(xa)n converges at x= a andn=0

anxn

converges at x= 0.


Remark.

A power series always converges at the center

i.e.,n=0

an(xa)n converges at x= a andn=0

anxn

converges at x= 0.So, our aim is to determine the values x, a for

whichn=0

an(xa)n converges.


The following result gives us all the information we

can have regarding convergence/divergence of a

given power series.




given power series.

Theorem (Theorem 18. The Convergence

Theorem for Power Series)

If the power seriesn=0

anxn

(i) Converges at x= c(c, 0), then it convergesabsolutely for all x such that |x| < |c|.




given power series.

Theorem (Theorem 18. The Convergence

Theorem for Power Series)

If the power seriesn=0

anxn

(i) Converges at x= c(c, 0), then it convergesabsolutely for all x such that |x| < |c|.

(ii) Diverges at x= d, then it diverges for all xsuch that |x| > |d|.


Remark.

A power seriesan(xa)n can have only one of the

following three features:


Remark.



(i) It converges only at x= a and divergeselsewhere.


Remark.




(ii) It converges over an interval i.e., there is a

positive real number R such that series

converges absolutely for |xa| R. The series may ormay not converge at either endpoints

x= aR and x= a+R.


Remark.




(ii) It converges over an interval i.e., there is a

positive real number R such that series

converges absolutely for |xa| R. The series may ormay not converge at either endpoints

x= aR and x= a+R.(iii) It converges absolutely for all x.Devendra Kumar BITS, Pilani Mathematics I

In case (ii), the interval is called the interval of

convergence of the power series (it may includes one

or both end points) and the half length of the interval

is called the radius of convergence of the power

series (we denote it by R).


Radius of Convergence

A nonnegative number R such that the power seriesn=0

an(xa)n converges absolutely for x such that|xa| R is called theradius of convergence of the power series. The series

may or may not converge at either endpoints

x= aR and x= a+R.


Formula for Radius of Convergence

For the power seriesn=0

an(xa)n the radius ofconvergence is given by the formulae





1

R= lim

an+1an .





1

R= lim

an+1an .

or1

R= lim |an|1/n .





1

R= lim

an+1an .

or1

R= lim |an|1/n .

Remark.

If series converges only at center then R = 0.





1

R= lim

an+1an .

or1

R= lim |an|1/n .

Remark.

If series converges only at center then R = 0.If series converges for all x then R =.


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