Section 11.5 – Testing for Convergence at Endpoints

Common Series to be used….

Harmonic Series - n 1

1 1 1 11 ...

n 2 3 4

DIVERGES

p-Seriesp p p p

n 1

1 1 1 11 ...

n 2 3 4

Converges if p > 1Diverges if p < 1

Comparison Test for Convergence

n n n nconverges conva if c and a c for alle ges nr

Comparison Test for Divergence

n n n ndiverges divea if d and a drges for all n

3n 1

n 1

n 3

nnlima 0

3

n 1

n 3

acts like 2

1as n

n

3n 1

n 1

n 3

converges by the comparison test

to the p-series with p = 2.

Makes terms smaller

Makes terms smaller

2

2n 1

n 1

n 3n 2

n

nlima 1

2

n 2nn 1

n 1Since lima 1,

n 3n 2

diverges by the nth term test

for divergence.

2

3n 1

6n 1

2n 1

n

nlima 0

2

3

6n 1

2n 1

acts like3

as nn

2

3n 1

6n 1

2n 1

diverges by the comparison test to

the p-series with p = 1.

n nIf a converges, then a converges

n

n

A series a is said to CONVERGE ABSOLUTELY

if a converges

ABSOLUTE CONVERGENCE

2n 1

1converges absolutely

n 1

by comparison to p-series with p=2>1.

2n 1

1Test for convergence.

n 2n 1

n 1

2n 1

11

n

2n 1

1Since is a p-series with p = 2>1, it converges.

n

n 1

2n 1

11 converges absolutely.

n

2n 1

1 n

n

2n 1

1Since converges (p-series with p = 2>1)

n

n 1

1 and diverges (harmonic series)

n

2n 1

1 ndiverges

n

n 1

cosn

n n

n 1

cosnconverges absolutely

n n

3

by comparison to p-series with p= >1.2

n

nn 1

n n1 diverges since lim 1 0

n 1 n 1

n

n 1

5

n

n 1

5 converges

-1

since infinite geometric series with r = 5

n

n 1

n1

n 1

n n 1

n n nn=1 n=1

If a 0, then an alternating series -1 a or -1 a

Alternating Series Test

converges if both of the following conditions are satisfied:

n

n

n n+1 n

1. lima 0

2. a is a decreasing sequence for all n, or a a

If a series converges but the series of absolute values diverges,We say the series converges conditionally.

CONDITIONAL CONVERGENCE – (AST)

Determine whether the series is converges conditionally, converges absolutely, or diverges.

n 1

cos n

n

n

11. lim 0

n

1 12.

n 1 n

The series converges conditionally

n 1

cosndoes not converge absolutely

n by comparison test to the p-series with p=1/2<1

Determine whether the series is converges conditionally, converges absolutely, or diverges.

n 1 5

n 1

1 n

n 1 5

n 1

1 n

by comparison test t

converges absol

o the p-series

utely

with p=5>1

Determine whether the series is converges conditionally, converges absolutely, or diverges.

n 1

n 1

n1

n 1

n

n1. lim 1

n 1

The series diverges by the nth term test for divergence

n 1 n n 1

nn 1 n 1 n 1

Which of the following series are conditionally convergent?

1 cosn 1I. 1 II. 1 III. 1

2n 1 3 nA. I only B. II only C. I and II only D. I and III only E. I, II, III

n 1

n 0n 1

1 1I. 1 1. lim 0

2n 1 2TRUE

n 1

1 12.

2 n 1 1 2n 1TRUE

n

n nn 0n 1

cosn cosnII. 1 1. lim 0

3 3TRUE 1

n 1

n 0n 1

1 1III. 1 1. lim 0

n n1 1

2.n

TRUE

TRU1

En

2 2 2

n

n

2

Which of the following series are convergent?

1 1 1I. 1 ... ...

2 3 n

11 1II. 1 ... ...

2 3 n

8 2III. 2 1 ... ...

9 nA. I only B. III only C. I and II only D. II and III only E. I, II, III

I. p-series with p CON= 2 VE >1 RGES

II. alt. harmonic CONser VEies RGES

2n 1 2

2 nn n

2 n nl

III. ratio

im lim 2 22 n 1n

DIVERG

1

E test S

Find the interval of convergence for n

n 1

x

n

n 1

nn

x nlim 1

n 1 x

n

nx lim 1

n 1

x 1

If x = -1, n

n 1

1

n

converges, alternating series test

If x = 1, n

n 1

1

n

diverges, harmonic series

Interval of convergence [-1, 1)

1 x 1 ? ?

Find the interval of convergence for n

nn 0

nx 3

4

n 1 n

n 1 nn

n 1 x 3 4lim 1

4 n x 3

n

n 1x 3 lim 1

4n

x 3 4

If x = -7, n

nn 1

n

4

4

diverges, nth term test

If x = 1, n

nn 1 n 14

4nn

diverges, nth term test

Interval of convergence (-7, 1)

n

n 1

1 n

7 x 1 ? ?

Find the interval of convergence for n n

n 0

2 x

n!

n 1 n 1

n nn

2 x n!lim 1

n 1 ! 2 x

n

2x lim 1

n 1

0 1

Interval of convergence ,

Find the interval of convergence for

n 1 n 1

n nn

1 x 3 n 1lim 1

n 2 1 x 3

n

n 1x 3 lim 1

n 2

x 3 1

If x = 2, n n

n 0

1

n

1

1

diverges, harmonic series

If x = 4, n n n

n 1 n 1

1 1

n 1 n 1

1

converges, alternating series test

Interval of convergence (2, 4]

n n

0

1 x 3

n 1

2 x 4 ? ?

5

1 1a

2 5 10

5 1

5

1 1a

3 729

5 3

1 1a

1255

Find an upper bound for the error if the sum of the first fourterms is used as an approximation to the sum of the series.

n 1

n 1

114.

2n

n 1

n 1

116.

3

n 1

3n 1

118.

n

n 1a 0.005

1 1

2 n 1 200

2 n 1 200

n 99

n 1a 0.005

n 2

1 1

2003

n 2003

9

n 3

Find the smallest value of n for which the nth partial sum approximates the sum of the series within 0.005.

n 1

n 1

114.

2n

n 1

n 1

116.

3

n 1

3n 1

118.

n

n 1a 0.005

3

1 1

200n 1

3n 1 200

n 5