Raduis of Convergence

8/12/2019 Raduis of Convergence

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9.4 Radius of Convergence

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2008 Abraham Lincolns Home

Springfield, Illinois


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The first requirement of convergence is that the termsmust approach zero.

n th term test for divergence

1 n

n

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

a

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

Ex. 2:0

! nn

n x If then grows without

bound.

1 x ! nn x

If then0 1 x 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)


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(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 endpointsof the interval.

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

2 The series converges for every x . ( ) R

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

x a

The number R is the radius of convergence.


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This series converges.

So this series mustalso converge.

Direct Comparison Test

For non-negative series:If every term of a series isless than the correspondingterm of a convergent series,then both series converge.

If every term of a series isgreater than thecorresponding term of adivergent series, then bothseries diverge.

So this series must also diverge.

This series diverges.


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

2

20 !

n

n

x

n

There are no negative terms: 2 2

2 !!

nn x x

nn

2

0 !

n

n

x

nis the Taylor series for , which converges.

2 x

e

larger denominator

The original series converges.

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


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Absolute Convergence

If converges, then we say converges absolutely.na na

The term converges absolutely means that the seriesformed by taking the absolute value of each termconverges. Sometimes in the English language we use

the word absolutely to mean really or actually. This isnot the case here!

If converges, then converges.na na

If the series formed by taking the absolute value of eachterm converges, then the original series must alsoconverge.

If a series converges absolutely, then it converges.


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Ex. 4: 0

sin

!

n

n

x

n

We test for absolute convergence:sin 1

! !

n x

n n

Since ,2 3

1

2! 3! !

n x x x xe x

n

0

1

!n n

converges to1e e

0

sin

!

n

n

x

nconverges by the direct comparison test.

Since converges absolutely, it converges.

0

sin

!

n

n

x

n


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Ratio Technique

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

1

1

1

n

n

r S t

r where r = common ratio between terms

When , the series converges.1r


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Geometric series have a constant ratio between terms.Other series have ratios that are not constant. If theabsolute value of the limit of the ratio betweenconsecutive terms is less than one, then the series willconverge.

For , if then:1 nnt 1lim n

nn

t L

t

if the series converges.1 L

if the series diverges.1 L

if the series may or may not converge.1 L


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2 3 4

ln 12 3 4

x x x x x

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

2 3 41 1 1

ln 1 1 1 12 3 4

x 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 n L

n x

1 1lim

1 1

n

nn

x x n

n x

1lim

1n

x n

n 1 x

If the limit of the ratiobetween consecutive termsis less than one, then the

series will converge.

11

1n

n

n n

aa

a a


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1 1 x

1 1 1 x

0 2 x

The interval of convergence is (0,2).

The radius of convergence is 1.

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


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Ex:

1

53

n

nn

n x

2 31 2 3

5 5 53 9 27

x x x

1

1

1 5 3lim

3 5

n n

nnn

n x L

n x

1 5 5 3lim

3 3 5

n n

nnn

n x x L

n x

1 5lim

3n

n x L

n


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Ex:

1

53

n

nn

n x

1 5lim3n

n x L

n

15 lim3n

n L xn

15

3 L x

15 1

3

x

5 3 x

3 5 3 x

2 8 x

The interval of convergence is (2,8).

The radius of convergence is .8 2

32


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Ex: 4

1

!3

n

n

n x

n

2 3 41 2 3

3 3 3 38 27 32

x x x x

1 4

4

1 ! 3lim

1 ! 3

n

nn

n x n L

n n x

4

4

! 1 3 3lim

1 ! 3

n

nn

n n x x n L

n n x

4

3 lim 1

1n

n L x n

n

1


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Ex: 4

1

!3

n

n

n x

n

4

3 lim 11n

n L x n

n

1

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

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

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


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Another series for which it is easy to find thesum 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 11n n n

1 1 1 11 1

2 3 3 42

3

11

4S

11

1n

S n

lim 1n

n

S

1

11

A B

n n nn

1 1 A n Bn

1 An A Bn

Telescoping Series

11

n n

n

b b

converges to 1 1lim nn

b b

p

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