M38 Lec 112813

MATH 38Mathematical Analysis III

I. F. Evidente

IMSP (UPLB)

Previously...

1 Convergence of an infinite series by definition2 Divergence Test3 Tests for Infinite Series of Nonnegative Terms:

1 Integral Test2 Limit Comparison Test

4 Special Series: geometric, harmonic, constant, p-series

Outline

1 Tests for Series of Nonnegative Terms (continued)Comparison Test

2 Test for Alternating Series

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.

Example

1n=k

1

n, where k is any integer greater than 1

(divergent)

2n=k

5

n7/5(convergent)

Example

1n=k

1

n, where k is any integer greater than 1 (divergent)

2n=k

5

n7/5(convergent)

Example

1n=k

1


2n=k

5

n7/5

(convergent)

Example

1n=k

1


2n=k

5

n7/5(convergent)

Outline



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

Remark

1 If 0< a < b, then 1a> 1

b

2 Given a positive fraction: If the numerator is held constant and1 the denominator is increased, the fraction will decrease2 the denominator is decreased, the fraction will increase

Remark

1 If 0< a < b, then 1a> 1

b2 Given a positive fraction: If the numerator is held constant and

1 the denominator is increased, the fraction will decrease2 the denominator is decreased, the fraction will increase

Examplen=1

5

n3n

n=1

5

n3n

n=1

5

3n

Examplen=1

|cosn|n2+1

n=1

|cosn|n2+1

n=1

1

n2

Examplen=2

1

lnn

n=2

1

lnn>

n=2

1

n>

The series is divergent.

Examplen=2

1

lnn

n=2

1

lnn>

n=2

1

n

>


Examplen=2

1

lnn

n=2

1

lnn>

n=2

1

n>


Examplen=1

(1+ 2

n

)n

n=1

(1+ 2

n

)n>

n=1

1>


Examplen=1

(1+ 2

n

)n

n=1

(1+ 2

n

)n>

n=1

1

>


Examplen=1

(1+ 2

n

)n

n=1

(1+ 2

n

)n>

n=1

1>


Outline



DefinitionAn alternating series is a series whose terms form an alternatingsequence.

(That is, the signs alternate from positive to negative.)

DefinitionAn alternating series is a series whose terms form an alternatingsequence. (That is, the signs alternate from positive to negative.)

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. (Recall two ways to show a sequenceis decreasing.)

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




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

(Recall two ways to show a sequenceis decreasing.)



n |an | = 02 {|an |} is a decreasing sequence. (Recall two ways to show a sequence

is decreasing.)

Examplen=1

(1)nn

(Alternating Harmonic Series)

|an | = (1)nn

= 1n1 lim

n1

n= 0

2 Is {|an |} decreasing?Let f (n)= 1

n

f (n)= 1n2

< 0

Yes, {|an |} decreasing. The series is convergent since it passes AST.

RemarkThe alternating harmonic series is convergent.

Examplen=1

(1)nn


|an |

= (1)nn

= 1n1 lim

n1

n= 0


n

f (n)= 1n2

< 0



Examplen=1

(1)nn


|an | = (1)nn

= 1n

1 limn

1

n= 0


n

f (n)= 1n2

< 0



Examplen=1

(1)nn


|an | = (1)nn

= 1n1 lim

n1

n

= 02 Is {|an |} decreasing?

Let f (n)= 1n

f (n)= 1n2

< 0



Examplen=1

(1)nn


|an | = (1)nn

= 1n1 lim

n1

n= 0


n

f (n)= 1n2

< 0



Examplen=1

(1)nn


|an | = (1)nn

= 1n1 lim

n1

n= 0

2 Is {|an |} decreasing?

Let f (n)= 1n

f (n)= 1n2

< 0



Examplen=1

(1)nn


|an | = (1)nn

= 1n1 lim

n1

n= 0


n

f (n)= 1n2

< 0



Examplen=1

(1)nn


|an | = (1)nn

= 1n1 lim

n1

n= 0


n

f (n)

= 1n2

< 0



Examplen=1

(1)nn


|an | = (1)nn

= 1n1 lim

n1

n= 0


n

f (n)= 1n2

< 0



Examplen=1

(1)nn


|an | = (1)nn

= 1n1 lim

n1

n= 0


n

f (n)= 1n2

< 0

Yes, {|an |} decreasing.

The series is convergent since it passes AST.


Examplen=1

(1)nn


|an | = (1)nn

= 1n1 lim

n1

n= 0


n

f (n)= 1n2

< 0



Examplen=1

(1)n1lnn

|an |

= (1)n1lnn

= 1lnn1 lim

n1

lnn= 0


lnn

f (n)= 1n(lnn)2

< 0


Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn

1 limn

1

lnn= 0


lnn

f (n)= 1n(lnn)2

< 0


Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn1 lim

n1

lnn

= 02 Is {|an |} decreasing?

Let f (n)= 1lnn

f (n)= 1n(lnn)2

< 0


Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn1 lim

n1

lnn= 0


lnn

f (n)= 1n(lnn)2

< 0


Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn1 lim

n1

lnn= 0

2 Is {|an |} decreasing?

Let f (n)= 1lnn

f (n)= 1n(lnn)2

< 0


Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn1 lim

n1

lnn= 0


lnn

f (n)= 1n(lnn)2

< 0


Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn1 lim

n1

lnn= 0


lnn

f (n)

= 1n(lnn)2

< 0


Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn1 lim

n1

lnn= 0


lnn

f (n)= 1n(lnn)2

< 0


Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn1 lim

n1

lnn= 0


lnn

f (n)= 1n(lnn)2

< 0

Yes, {|an |} decreasing.

The series is convergent since it passes AST.

Examplen=1

(1)n1lnn

|an | = (1)n1lnn

= 1lnn1 lim

n1

lnn= 0


lnn

f (n)= 1n(lnn)2

< 0


Remark1 Please use this only for an alternating series. :)

2 If an alternating series does not satisfy the above conditions, youcannot conclude that the alternating series is divergent.

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

Remark1 Please use this only for an alternating series. :)2 If an alternating series does not satisfy the above conditions, you

cannot conclude that the alternating series is divergent.

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

Remark1 Please use this only for an alternating series. :)2 If an alternating series does not satisfy the above conditions, you

cannot conclude that the alternating series is divergent.3 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). 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



The End. Thank you.

Tests for Series of Nonnegative Terms (continued)Comparison Test

Test for Alternating Series

M38 Lec 112813

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Transcript of M38 Lec 112813