M38 Lec 112613

MATH 38Mathematical Analysis III

I. F. Evidente

IMSP (UPLB)

Previously...

1 Infinite Series2 Definition of convergence of infinite series3 Sum of a convergent infinite series4 Special Series

1 Geometric Series2 Harmonic Series

Outline

1 The Divergence Test

2 Convergence Tests for Infinite Series of Nonnegative TermsThe Integral TestLimit Comparison TestComparison Test

Theorem (The Divergence Test)If lim

nan 6= 0, then the seriesan diverges.

ExampleDetermine whether the series converges or diverges.

1n=1

n

n+12

n=2

ln

(2n

n1)

3n=0

5

Remark (Not in handout)

The constant seriesn=k

c diverges if c 6= 0 and converges if c = 0.


1n=1

n

n+12

n=2

ln

(2n

n1)

3n=0

5



c diverges if

c 6= 0 and converges if c = 0.


1n=1

n

n+12

n=2

ln

(2n

n1)

3n=0

5



c diverges if c 6= 0 and converges if

c = 0.


1n=1

n

n+12

n=2

ln

(2n

n1)

3n=0

5



c diverges if c 6= 0 and converges if c = 0.

RemarkThe divergence test is one of the easiest ways to show that a series isdivergent!

Remark (Contrapositive of Divergence Test)Equivalently: If

an converges, then lim

nan = 0.

Notelimnan = 0 is a NECESSARY condition for convergence.

Example

Determine whethern=1

n2+15n3+1 converges or diverges.

Remarklimnan = 0 DOES NOT imply that

an converges. If lim

nan = 0, thenNO CONCLUSION can be made. That is,

an may converge or diverge.

Notelimnan = 0 is NOT A SUFFICIENT condition for convergence.

Example




an converges.

If limnan = 0, then

NO CONCLUSION can be made. That is,an may converge or diverge.


Example





nan = 0, thenNO CONCLUSION can be made.

That is,an may converge or diverge.


Example





nan = 0, thenNO CONCLUSION can be made. That is,

an may converge or diverge.


Question

Can you give an example of a special seriesn=1

an for which limnan = 0,

butn=1

an is divergent?

n=1

1

n

Outline

1 The Divergence Test

2 Convergence Tests for Infinite Series of Nonnegative TermsThe Integral TestLimit Comparison TestComparison Test

RemarkMake sure that all test assumptions are satisfied

Test assumptions need to be true only for the tail end of the seriesTest assumptions need NOT be true for a finite number of terms atthe beginning of the series

Yes, a googol (10100) is still finite.

Caveat to all those referring to the handout from other classes. :)

RemarkMake sure that all test assumptions are satisfiedTest assumptions need to be true only for the tail end of the series

Test assumptions need NOT be true for a finite number of terms atthe beginning of the series



RemarkMake sure that all test assumptions are satisfiedTest assumptions need to be true only for the tail end of the seriesTest assumptions need NOT be true for a finite number of terms atthe beginning of the series



Theorem (Integral Test)

Supposen=k

an is a series with positive terms.

Let f (n)= an . Considerf (x) as a function of a real variable. If f (x) is continuous, decreasing andpositive for all real x [k,), then

n=k

an is convergent if and only if k

f (x)dx is convergent.

(That is, the series converges if the the improper integral converges anddiverges is the improper integral diverges.)


Supposen=k

an is a series with positive terms. Let f (n)= an .

Consider

f (x) as a function of a real variable. If f (x) is continuous, decreasing andpositive for all real x [k,), then

n=k





Supposen=k

an is a series with positive terms. Let f (n)= an . Considerf (x) as a function of a real variable.

If f (x) is continuous, decreasing andpositive for all real x [k,), then

n=k





Supposen=k

an is a series with positive terms. Let f (n)= an . Considerf (x) as a function of a real variable. If f (x) is continuous, decreasing andpositive for all real x [k,), then

n=k




ExampleCheck assumptions!!!

1n=1

1

n2

2n=1

1pn

3n=2

1

n (lnn)3/2

TIPThe integral test is useful when an = f (n) is easy to integrate.

Definition (p-series or Hyperharmonic series)n=1

1

np(p > 0)

NoteThe harmonic series is a p-series where p = 1.

Theorem (Convergence of p-series)n=1

1

npconverges if p > 1 and diverges if 0< p 1.

Theorem (Two of the Four Convergence Theorems)Suppose

an and

bn are infinite series and c is a nonzero constant.

1 Ifan converges, then

can converges and

can = can . If an

diverges, thencan diverges.

2 Ifan and

bn differ only by a finite number of terms, then either

both converge or both diverge. (That is, convergence or divergence isunaffected by deleting a finite number of terms from the series.)

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent

.

2n=1

pi

n2is convergent since

n=1

1

n2is convergent.

3n=2

1

n= 12+ 13+ 14+ ... differs from harmonic series by one term. Thus,

it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ... differsn=1

1

n2by 9 terms. Thus, it

is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1


it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ... differsn=1

1


is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1

n

= 12+ 13+ 14+ ... differs from harmonic series by one term. Thus,

it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ... differsn=1

1


is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1

n= 12+ 13+ 14+ ...

differs from harmonic series by one term. Thus,

it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ... differsn=1

1


is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1

n= 12+ 13+ 14+ ... differs from harmonic series by one term.

Thus,

it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ... differsn=1

1


is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1


it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ... differsn=1

1


is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1


it is divergent.

4n=0

1

(n+10)2

= 1102

+ 1112

+ 1122

+ ... differsn=1

1


is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1


it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ...

differsn=1

1


is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1


it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ... differsn=1

1

n2by 9 terms.

Thus, it

is convergent.

Example

1n=1

p2

nis divergent since

n=1

1

nis divergent .

2n=1

pi


n=1

1

n2is convergent.

3n=2

1


it is divergent.

4n=0

1

(n+10)2 =1

102+ 1112

+ 1122

+ ... differsn=1

1


is convergent.

Theorem (Limit Comparison Test)Letan and

bn be infinite series of positive terms.

1 If limn

anbn

= L > 0, then

the two series either both converge or both

diverge.

2 If limn

anbn

= 0 and bn is convergent, then an is convergent.3 If lim

nanbn

= and bn is divergent, then an is divergent.



1 If limn

anbn

= L > 0, then the two series either both converge or bothdiverge.

2 If limn

anbn


nanbn




1 If limn

anbn


2 If limn

anbn

= 0 and bn is convergent, then

an is convergent.

3 If limn

anbn




1 If limn

anbn


2 If limn

anbn

= 0 and bn is convergent, then an is convergent.

3 If limn

anbn




1 If limn

anbn


2 If limn

anbn


nanbn

= and bn is divergent, then

an is divergent.



1 If limn

anbn


2 If limn

anbn


nanbn


Remark

Put the nth TERM of the series you are testing in the numerator, and thenth TERM of the special series you are comparing it to in thedenominator.

NoteThe limit comparison test is useful when f (n) is a rational function oralgebraic function:

Not in handout: If f (n)= g (n)h(n)

, compare the series with

n=1

1

np

where p = degree of hdegree of g .

Example

1n=1

n21n5

2n=1

p2n13n+1

3n=2

1

ln(lnn)

(Dapat terms, hindi series sa quotient)

Theorem (Comparison Test)Letan and

bn be infinite series of nonnegative terms such that an bn

for all n = 1,2, ...

1 Ifbn converges, then

an converges.

2 Ifan diverges, then

bn diverges.

Theorem (Comparison Test)Letan and

bn be infinite series of nonnegative terms such that an bn

for all n = 1,2, ...1 If

bn converges, then

an converges.

2 Ifan diverges, then

bn diverges.

RemarkDifficulty with comparison test: you have to make an intelligent guess as towhether or not the series is convergent.

If your guess is convergent, try to find a "greater" convergent specialseries.If your guess divergent, try to find a "lesser" divergent special series.


If your guess is convergent,

try to find a "greater" convergent specialseries.If your guess divergent, try to find a "lesser" divergent special series.


If your guess is convergent, try to find a "greater" convergent specialseries.

If your guess divergent, try to find a "lesser" divergent special series.


If your guess is convergent, try to find a "greater" convergent specialseries.If your guess divergent,

try to find a "lesser" divergent special series.


If your guess is convergent, try to find a "greater" convergent specialseries.If your guess divergent, try to find a "lesser" divergent special series.

With some abuse of notation :)

TipConvergent:

an

bn

TipThe comparison test could be useful if f (n) involves sin, cos, ln or Arctanbecause:

1 sinn 1; 1 cosn 10 lnn pn n if n 1pi2

y = x

1

y =pxy = lnx

Example

1n=1

5

n3n

2n=1

|cosn|n2+1

3n=2

1

lnn

4n=1

lnn

n2

In Summary

1 What is the easiest way to prove that a series is divergent?

2 Givenn=1

an such that limnan = 0, what can you conclude about the

series?3 What are the three special series taken up today?

1 What can you say about the convergence of the constant series?2 What can you say about the convergence of the p-series?3 What can you say about the convergence of the alternating harmonic

series?4 What are the three tests for series of nonnegative terms? When is it

best to use1 integral test2 limit comparison test3 comparison test

In Summary


2 Givenn=1


series?

3 What are the three special series taken up today?1 What can you say about the convergence of the constant series?2 What can you say about the convergence of the p-series?3 What can you say about the convergence of the alternating harmonic



In Summary


2 Givenn=1






In Summary


2 Givenn=1



1 What can you say about the convergence of the constant series?

2 What can you say about the convergence of the p-series?3 What can you say about the convergence of the alternating harmonic



In Summary


2 Givenn=1



1 What can you say about the convergence of the constant series?2 What can you say about the convergence of the p-series?

3 What can you say about the convergence of the alternating harmonicseries?

4 What are the three tests for series of nonnegative terms? When is itbest to use

1 integral test2 limit comparison test3 comparison test

In Summary


2 Givenn=1




series?

4 What are the three tests for series of nonnegative terms? When is itbest to use


In Summary


2 Givenn=1




series?4 What are the three tests for series of nonnegative terms?

When is itbest to use


In Summary


2 Givenn=1





best to use1 integral test

2 limit comparison test3 comparison test

In Summary


2 Givenn=1





best to use1 integral test2 limit comparison test

3 comparison test

In Summary


2 Givenn=1






Homework

Given the (controversial) infinite seriesn=1

(1)n+1 = 11+11+1 ...

1 SPS: {1,0,1,0, ...}2 The infinite series is divergent; we do not speak of its sum.

Euler said:Notable enough, however, are the controversies over the series 1 - 1 + 1 -1 + 1 - ... whose sum was given by Leibniz as 1/2, although othersdisagree. ... The sum of a series...has relevance only for convergent series,and we should in general give up the idea of sum for divergent series

Remarkn=1

(1)n+1 is a geometric series with r =1. Ergo, it is divergent.

Homework


(1)n+1 = 11+11+1 ...1 SPS: {1,0,1,0, ...}

2 The infinite series is divergent; we do not speak of its sum.


Remarkn=1


Homework


(1)n+1 = 11+11+1 ...1 SPS: {1,0,1,0, ...}2 The infinite series is divergent;

we do not speak of its sum.


Remarkn=1


Homework


(1)n+1 = 11+11+1 ...1 SPS: {1,0,1,0, ...}2 The infinite series is divergent; we do not speak of its sum.


Remarkn=1


The End. Thank you.

The Divergence TestConvergence Tests for Infinite Series of Nonnegative TermsThe Integral TestLimit Comparison TestComparison Test

M38 Lec 112613

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