M38 Lec 120313

MATH 38Mathematical Analysis III

I. F. Evidente

IMSP (UPLB)

Previously...

1 Tests for Infinite Series of Nonnegative Terms:1 Integral Test2 Limit Comparison Test3 Comparison Test

2 Alternating Series Test3 Special Series: geometric, harmonic, constant, p-series, alternating

harmonic series

Outline

1 Four Convergence Theorems

2 Absolute and Conditional Convergence

3 Absolute Convergence TestsRoot TestRatio Test

Theorem (Alternating Series Test for Convergence)Letan be an alternating series.

The alternating series is convergent ifboth of the following conditions hold:

1 limn |an | = 0

2 {|an |} is a decreasing sequence.

Theorem (Alternating Series Test for Convergence)Letan be an alternating series. The alternating series is convergent if

both of the following conditions hold:

1 limn |an | = 0



both of the following conditions hold:1 lim

n |an | = 0



both of the following conditions hold:1 lim

n |an | = 02 {|an |} is a decreasing sequence.

Remark1 If an alternating series does not satisfy the above conditions, you

cannot conclude that the alternating series is divergent.

2 If the first condition is not satisfied, use the divergence test to provethat the sequence is divergent.

Remark1 If an alternating series does not satisfy the above conditions, you

cannot conclude that the alternating series is divergent.2 If the first condition is not satisfied, use the divergence test to prove

that the sequence is divergent.

Examplen=1

(1)nnn+1

|an | = (1)nnn+1

= nn+1limn

n

n+1 = 1 limnn

n+1 =1

This implies: limn

(1)nnn+1 does not exist (see remark after Thm 1.1.3).

The series is divergent by the Divergence Test.

Examplen=1

(1)nnn+1

|an |

= (1)nnn+1

= nn+1limn

n

n+1 = 1 limnn

n+1 =1

This implies: limn



Examplen=1

(1)nnn+1

|an | = (1)nnn+1

= nn+1

limn

n

n+1 = 1 limnn

n+1 =1

This implies: limn



Examplen=1

(1)nnn+1

|an | = (1)nnn+1

= nn+1limn

n

n+1

= 1 limn

n

n+1 =1

This implies: limn



Examplen=1

(1)nnn+1

|an | = (1)nnn+1

= nn+1limn

n

n+1 = 1

limn

n

n+1 =1

This implies: limn



Examplen=1

(1)nnn+1

|an | = (1)nnn+1

= nn+1limn

n

n+1 = 1 limnn

n+1

=1

This implies: limn



Examplen=1

(1)nnn+1

|an | = (1)nnn+1

= nn+1limn

n

n+1 = 1 limnn

n+1 =1

This implies: limn



Examplen=1

(1)nnn+1

|an | = (1)nnn+1

= nn+1limn

n

n+1 = 1 limnn

n+1 =1

This implies: limn

(1)nnn+1

does not exist (see remark after Thm 1.1.3).


Examplen=1

(1)nnn+1

|an | = (1)nnn+1

= nn+1limn

n

n+1 = 1 limnn

n+1 =1

This implies: limn



Outline




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.

Simply put:1 Convergence or divergence of a series is unaffected by multiplying the

terms of the series by a constant.2 Convergence or divergence of a series is unaffected by deleting or

adding a finite number of terms.


an and



can converges and

can = can .

Ifan


2 Ifan and







an and



can converges and

can = can . If an

diverges, then

can diverges.

2 Ifan and







an and



can converges and

can = can . If an


2 Ifan and







an and



can converges and

can = can . If an


2 Ifan and



Simply put:

1 Convergence or divergence of a series is unaffected by multiplying theterms of the series by a constant.

2 Convergence or divergence of a series is unaffected by deleting oradding a finite number of terms.


an and



can converges and

can = can . If an


2 Ifan and




terms of the series by a constant.

2 Convergence or divergence of a series is unaffected by deleting oradding a finite number of terms.


an and



can converges and

can = can . If an


2 Ifan and






Theorem (Additional Two of the Four Convergence Theorems)1 Suppose

an and

bn both converge. Then

(an +bn) and

(an bn) both converge and (an +bn)=an +bn and(an bn)=an bn

2 If one series diverges and the other converges, then(an +bn)

diverges.


an and


(an +bn) and

(an bn) both converge and

(an +bn)=an +bn and

(an bn)=an bn2 If one series diverges and the other converges, then

(an +bn)

diverges.


an and


(an +bn) and



diverges.


an and


(an +bn) and


2 If one series diverges and the other converges, then

(an +bn)

diverges.


an and


(an +bn) and



diverges.

ExampleDetermine whether or not the following series converge:

1n=1

(1

2n 2

n+1

3n

)

(convergent)

2n=1

(5

n+ (1)

n

n

)(divergent)

3 What is the sum of the series in (1)?

RemarkIfan and

bn both diverge, then nothing can be said about the

convergence of(an +bn).


1n=1

(1

2n 2

n+1

3n

)(convergent)

2n=1

(5

n+ (1)

n

n

)(divergent)


RemarkIfan and




1n=1

(1

2n 2

n+1

3n

)(convergent)

2n=1

(5

n+ (1)

n

n

)

(divergent)


RemarkIfan and




1n=1

(1

2n 2

n+1

3n

)(convergent)

2n=1

(5

n+ (1)

n

n

)(divergent)


RemarkIfan and



Outline




Motivation:Given

an , we consider the convergence of

|an |.

DefinitionLetan be an infinite series.

1an is said to be absolutely convergent if

|an | is convergent.2an is said to be conditionally convergent if

an is convergent,

but |an | is divergent.

NoteWalang conditionally divergent :)

Motivation:Given


|an |.DefinitionLetan be an infinite series.



an is convergent,



Motivation:Given




|an | is convergent.

2an is said to be conditionally convergent if

an is convergent,



Motivation:Given





an is convergent,



Example

1n=1

(1)n+12n

n=1

(1)n+12n =

n=1

1

2n(convergent)

Thus,n=1

(1)n+12n

is absolutely convergent.

2n=1

(1)nn

n=1

(1)nn =

n=1

1

n(divergent)

Thus,n=1

(1)nn

is conditionally convergent.

Example

1n=1

(1)n+12n

n=1

(1)n+12n

=n=1

1

2n(convergent)

Thus,n=1

(1)n+12n


2n=1

(1)nn

n=1

(1)nn =

n=1

1

n(divergent)

Thus,n=1

(1)nn


Example

1n=1

(1)n+12n

n=1

(1)n+12n =

n=1

1

2n

(convergent)

Thus,n=1

(1)n+12n


2n=1

(1)nn

n=1

(1)nn =

n=1

1

n(divergent)

Thus,n=1

(1)nn


Example

1n=1

(1)n+12n

n=1

(1)n+12n =

n=1

1

2n(convergent)

Thus,n=1

(1)n+12n


2n=1

(1)nn

n=1

(1)nn =

n=1

1

n(divergent)

Thus,n=1

(1)nn


Example

1n=1

(1)n+12n

n=1

(1)n+12n =

n=1

1

2n(convergent)

Thus,n=1

(1)n+12n


2n=1

(1)nn

n=1

(1)nn

=n=1

1

n(divergent)

Thus,n=1

(1)nn


Example

1n=1

(1)n+12n

n=1

(1)n+12n =

n=1

1

2n(convergent)

Thus,n=1

(1)n+12n


2n=1

(1)nn

n=1

(1)nn =

n=1

1

n

(divergent)

Thus,n=1

(1)nn


Example

1n=1

(1)n+12n

n=1

(1)n+12n =

n=1

1

2n(convergent)

Thus,n=1

(1)n+12n


2n=1

(1)nn

n=1

(1)nn =

n=1

1

n(divergent)

Thus,n=1

(1)nn





an is convergent,


Question1 Is a conditionally convergent series convergent?2 Is an absolutely convergent series convergent?




an is convergent,


Question1 Is a conditionally convergent series convergent?

2 Is an absolutely convergent series convergent?




an is convergent,


Question1 Is a conditionally convergent series convergent?2 Is an absolutely convergent series convergent?

TheoremIfan is absolutely convergent, then

an is convergent.

Let {an}n=1 be a sequence.

a1+a2+ ...+ak |a1+a2+ ...+ak | |a1|+ |a2|+ ...|ak |

kn=1

an k

n=1|an |

limk

kn=1

ak limk

kn=1

|ak |n=1

an n=1

|an |


an is convergent.


a1+a2+ ...+ak

|a1+a2+ ...+ak | |a1|+ |a2|+ ...|ak |

kn=1

an k

n=1|an |

limk

kn=1

ak limk

kn=1

|ak |n=1

an n=1

|an |


an is convergent.


a1+a2+ ...+ak |a1+a2+ ...+ak |

|a1|+ |a2|+ ...|ak |

kn=1

an k

n=1|an |

limk

kn=1

ak limk

kn=1

|ak |n=1

an n=1

|an |


an is convergent.


a1+a2+ ...+ak |a1+a2+ ...+ak | |a1|+ |a2|+ ...|ak |

kn=1

an k

n=1|an |

limk

kn=1

ak limk

kn=1

|ak |n=1

an n=1

|an |


an is convergent.


a1+a2+ ...+ak |a1+a2+ ...+ak | |a1|+ |a2|+ ...|ak |

kn=1

an k

n=1|an |

limk

kn=1

ak limk

kn=1

|ak |

n=1

an n=1

|an |


an is convergent.


a1+a2+ ...+ak |a1+a2+ ...+ak | |a1|+ |a2|+ ...|ak |

kn=1

an k

n=1|an |

limk

kn=1

ak limk

kn=1

|ak |n=1

an n=1

|an |

ProblemDraw a Venn diagram illustrating the relationship between the set ofconvergent series, the set of absolutely convergent series, the set ofconditionally convergent series and the set of divergent series.

CONVERGENT DIVERGENT

absolutelyconvergent

conditionallyconvergent

Outline




Theorem (Root Test)Letan be an infinite series.

1 If limn

n|an | = L < 1, then

the series is absolutely convergent.

2 If limn

n|an | = L > 1 or lim

nn|an | =, then the series is divergent.

3 If limn

n|an | = 1, then no conclusion can be made.

(Commit to memory)

Note1 np|an | = |an |1/n

2 (1) Apply absolute value, (2) Get the nth root, (3) Get the limit.


1 If limn

n|an | = L < 1, then the series is absolutely convergent.

2 If limn



3 If limn


(Commit to memory)




1 If limn


2 If limn


nn|an | =, then

the series is divergent.

3 If limn


(Commit to memory)




1 If limn


2 If limn



3 If limn


(Commit to memory)




1 If limn


2 If limn



3 If limn

n|an | = 1, then

no conclusion can be made.

(Commit to memory)




1 If limn


2 If limn



3 If limn


(Commit to memory)



TIP:The Root Test is useful when the index of summation is a power that canbe "extracted".


1 If limn


2 If limn



3 If limn


Example

1n=1

1

nn

(absolutely convergent)

2n=2

(1)n(2n

n+1)n

(divergent)

3n=1

2n+1

n2n(absolutely convergent)


1 If limn


2 If limn



3 If limn


Example

1n=1

1

nn(absolutely convergent)

2n=2

(1)n(2n

n+1)n

(divergent)

3n=1

2n+1



1 If limn


2 If limn



3 If limn


Example

1n=1

1


2n=2

(1)n(2n

n+1)n

(divergent)

3n=1

2n+1

n2n



1 If limn


2 If limn



3 If limn


Example

1n=1

1


2n=2

(1)n(2n

n+1)n

(divergent)

3n=1

2n+1


Theorem (Ratio Test)Letan be an infinite series for which each term is nonzero.

1 If limn

an+1an= L < 1, then the series is absolutely convergent.

2 If limn

an+1an= L > 1 or limn

an+1an=, then the series is divergent.

3 If limn

an+1an= 1, then no conclusion can be made.

(Commit to memory)

Note

1

an+1an= |an+1||a1n |

2 (1) Apply the absolute value, (2) Simplify, (3) Get the limit.


1 If limn

an+1an= L < 1, then

the series is absolutely convergent.

2 If limn



3 If limn


(Commit to memory)

Note

1




1 If limn


2 If limn



3 If limn


(Commit to memory)

Note

1




1 If limn


2 If limn


an+1an=, then

the series is divergent.

3 If limn


(Commit to memory)

Note

1




1 If limn


2 If limn



3 If limn


(Commit to memory)

Note

1




1 If limn


2 If limn



3 If limn

an+1an= 1, then

no conclusion can be made.

(Commit to memory)

Note

1




1 If limn


2 If limn



3 If limn


(Commit to memory)

Note

1



TIP:The Ratio Test is useful when the series involves factorials, or the index ofsummation is a power that cannot be "extracted".

Example

1n=1

1

n!


2n=1

(1)n+13n2n2+3 (divergent)

3n=1

n5

(2n)!(absolutely convergent)

Example

1n=1

1

n!(absolutely convergent)

2n=1


3n=1

n5


Example

1n=1

1


2n=1

(1)n+13n2n2+3

(divergent)

3n=1

n5


Example

1n=1

1


2n=1


3n=1

n5


Example

1n=1

1


2n=1


3n=1

n5

(2n)!


Example

1n=1

1


2n=1


3n=1

n5


RemarkTwo additional special series:

1n=1

1

n!is convergent.

2n=1

1

nnis convergent.

These may come in handy...

RemarkTo change the lower index of an infinite series:

n=k

f (n)=

n=k+1

f (n1)=

n=k1f (n+1)

Example

1n=0

1

n+1 =n=1

1

n

2n=1

ar n1 =n=0

ar n



n=k

f (n)=

n=k+1f (n1)=

n=k1

f (n+1)

Example

1n=0

1

n+1 =n=1

1

n

2n=1

ar n1 =n=0

ar n



n=k

f (n)=

n=k+1f (n1)=

n=k1

f (n+1)

Example

1n=0

1

n+1 =

n=1

1

n

2n=1

ar n1 =n=0

ar n



n=k

f (n)=

n=k+1f (n1)=

n=k1

f (n+1)

Example

1n=0

1

n+1 =n=1

1

n

2n=1

ar n1 =n=0

ar n



n=k

f (n)=

n=k+1f (n1)=

n=k1

f (n+1)

Example

1n=0

1

n+1 =n=1

1

n

2n=1

ar n1 =

n=0

ar n



n=k

f (n)=

n=k+1f (n1)=

n=k1

f (n+1)

Example

1n=0

1

n+1 =n=1

1

n

2n=1

ar n1 =n=0

ar n



n=k

f (n)=

n=k+1f (n1)=

n=k1

f (n+1)

Example

1n=0

1

n+1 =n=1

1

n

2n=1

ar n1 =n=0

ar

n



n=k

f (n)=

n=k+1f (n1)=

n=k1

f (n+1)

Example

1n=0

1

n+1 =n=1

1

n

2n=1

ar n1 =n=0

ar n


RemarkTo find the common ratio and first term of a geometric series, put it in theform

n=0ar n

Example

1n=0

2n+1

3n

2n=1

1

2n

3n=1

(1)n+15n13n2n+1

Proof: Convergence of Geometric Series

Caser > 0 and r 6= 1

Note: If 0< r < 1limnr

n+1 =

0

If r > 1limnr

n+1 =




n+1 = 0

If r > 1limnr

n+1 =




n+1 = 0If r > 1

limnr

n+1 =


Givenn=0

ar n such that r > 0 and r 6= 1.

The nth partial sum of is

sn = a(1 rn)

1 rWhen does the SPS converge?

limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)

If 0< r < 1:limn sn =

a

1 r limn(1 rn)= a

1 r The series converges in this case.If r > 1:

limn sn =

a

1 r limn(1 rn)=

The series diverges in this case.


Givenn=0

ar n such that r > 0 and r 6= 1. The nth partial sum of is

sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a


limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)

1 r

When does the SPS converge?

limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a


limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a


limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn =

limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a


limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a


limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)

If 0< r < 1:

limn sn =

a

1 r limn(1 rn)= a


limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a


limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)=

a


limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a

1 r

The series converges in this case.If r > 1:

limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a

1 r The series converges in this case.

If r > 1:limn sn =

a

1 r limn(1 rn)=



Givenn=0


sn = a(1 rn)


limn sn = limn

a(1 r n)1 r =

a

1 r limn(1 rn)


a

1 r limn(1 rn)= a


limn sn =

a

1 r limn(1 rn)=


Summary

1 What are the four convergence theorems?

2 What is the difference between general convergence, absoluteconvergence and conditional convergence?

3 What the two absolute convergence tests were taken up today?1 When is it useful to use the root test?2 When is it useful to use the ratio test?

Summary

1 What are the four convergence theorems?2 What is the difference between general convergence, absolute

convergence and conditional convergence?

3 What the two absolute convergence tests were taken up today?1 When is it useful to use the root test?2 When is it useful to use the ratio test?

Summary


convergence and conditional convergence?3 What the two absolute convergence tests were taken up today?

1 When is it useful to use the root test?2 When is it useful to use the ratio test?

Summary



1 When is it useful to use the root test?

2 When is it useful to use the ratio test?

Summary



1 When is it useful to use the root test?2 When is it useful to use the ratio test?

END OF CHAPTER 1 :)

Announcement:Chapter 1 Quiz will be on December 12, Thursday (Lecture Class)

END OF CHAPTER 1 :)

Announcement:Chapter 1 Quiz will be on December 12, Thursday (Lecture Class)

Four Convergence TheoremsAbsolute and Conditional ConvergenceAbsolute Convergence TestsRoot TestRatio Test

M38 Lec 120313

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