9.4 Radius of Convergence

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2008

Abraham Lincoln’s HomeSpringfield, Illinois

Convergence

The series that are of the most interest to us are those that converge.

Today we will consider the question:

“Does this series converge, and if so, for what values of x does it converge?”

The first requirement of convergence is that the terms must approach zero.

nth term test for divergence

1 nna

diverges if fails to exist or is not zero.lim nn

a

Note that this can prove that a series diverges, but can not prove that a series converges.

Ex. 2:0

! n

nn x

If then grows without

bound.

1x ! nn x

If then0 1x 1

!lim ! limn

nn n

x

nn x

As , eventually is larger than , therefore

the numerator grows faster than the denominator.

n n1

x The series diverges. (except when x=0)

(As in the previous example.)

There are three possibilities for power series convergence.

1 The series converges over some finite interval:(the interval of convergence).

The series may or may not converge at the endpoints of the interval.

There is a positive number R such that the series diverges for but converges for .x a R x a R

2 The series converges for every x. ( )R

3 The series converges at and diverges everywhere else. ( )0R

x a

The number R is the radius of convergence.

This series converges.

So this series must also converge.

Direct Comparison Test

For non-negative series:

If every term of a series is less than the corresponding term of a convergent series, then both series converge.

If every term of a series is greater than the corresponding term of a divergent series, then both series diverge.

So this series must also diverge.

This series diverges.

Ex. 3: Prove that converges for all real x.

2

20 !

n

n

x

n

There are no negative terms:

2 2

2 !!

nnx xnn

2

0 !

n

n

xn

is the Taylor series for , which converges.

2xe

larger denominator

The original series converges.

The direct comparison test only works when the terms are non-negative.

Absolute Convergence

If converges, then we say converges absolutely.na naThe term “converges absolutely” means that the series formed by taking the absolute value of each term converges. Sometimes in the English language we use the word “absolutely” to mean “really” or “actually”. This is not the case here!

If converges, then converges.na na

If the series formed by taking the absolute value of each term converges, then the original series must also converge.

“If a series converges absolutely, then it converges.”

Ex. 4: 0

sin

!

n

n

x

n

We test for absolute convergence:sin 1

! !

nx

n n

Since ,2 3

1 2! 3! !

nx x x x

e xn

0

1

!n n

converges to

1e e

0

sin

!

n

n

x

n

converges by the direct comparison test.

Since converges absolutely, it converges.

0

sin

!

n

n

x

n

Ratio Technique

We have learned that the partial sum of a geometric series is given by:

1

1

1

n

n

rS t

r

where r = common ratio between terms

When , the series converges.1r

Geometric series have a constant ratio between terms. Other series have ratios that are not constant. If the absolute value of the limit of the ratio between consecutive terms is less than one, then the series will converge.

For , if then:1

nn

t

1lim n

nn

tL

t

if the series converges.1L

if the series diverges.1L

if the series may or may not converge.1L

2 3 4

ln 12 3 4

x x xx x

Ex:If we replace x with x-1, we get:

2 3 41 1 1ln 1 1 1 1

2 3 4x x x x x 1

1

11 1

n n

n

xn

2 1

1

1 1lim

1 1 1

n n

n nn

x nL

n x

1 1lim

1 1

n

nn

x x n

n x

1lim

1n

x n

n

1x

If the limit of the ratio between consecutive terms is less than one, then the series will converge.

11

1nn

n n

aa

a a

1 1x

1 1 1x

0 2x

The interval of convergence is (0,2).

The radius of convergence is 1.

If the limit of the ratio between consecutive terms is less than one, then the series will converge.

Ex:

1

53

n

nn

nx

2 31 2 35 5 5

3 9 27x x x

1

1

1 5 3lim

3 5

n n

nnn

n xL

n x

1 5 5 3lim

3 3 5

n n

nnn

n x xL

n x

1 5lim

3n

n xL

n

Ex:

1

53

n

nn

nx

1 5lim

3n

n xL

n

15 lim

3n

nL x

n

15

3L x

15 1

3x

5 3x

3 5 3x

2 8x

The interval of convergence is (2,8).

The radius of convergence is .8 2

32

Ex: 4

1

!3

n

n

nx

n

2 3 41 2 33 3 3 3

8 27 32x x x x

1 4

4

1 ! 3lim

1 ! 3

n

nn

n x nL

n n x

4

4

! 1 3 3lim

1 ! 3

n

nn

n n x x nL

n n x

4

3 lim 11n

nL x n

n

1

Ex: 4

1

!3

n

n

nx

n

4

3 lim 11n

nL x n

n

1

L for all .3x Radius of convergence = 0.

At , the series is , which converges to zero.3x 0 0 0

Note: If R is infinite, then the series converges for all values of x.

Another series for which it is easy to find the sum is the telescoping series.

Ex. 6: 1

1

1n n n

Using partial fractions:

1 A 0 A B

0 1 B

1 B

1

1 1

1n n n

1 1 1 11 1

2 3 3 42

3

11

4S

11

1nSn

lim 1nnS

1

11

A B

n n nn

1 1A n Bn

1 An A Bn

Telescoping Series

11

n nn

b b

converges to 1 1lim nnb b