Chapter 11.9, Problem 39E

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

Chapter
Section

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

# Let f ( x ) = ∑ n = 1 ∞ x n n 2 Find the intervals of convergence for f, f′, and f″.

To determine

To find: The interval of convergence for f(x) , f(x) and f(x) .

Explanation

Given:

The series is f(x)=n=1xnn2 .

Result used:

(1)Ratio test:

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

(2) The p-series n=11n is converges if p>1 and diverges if p1 .

(3) Test for divergence: if limnan does not exist or limnan0 , then the series n=1an is divergent.

(4) Alternating Series Test:

“If the alternating series n=1(1)n1bn=b1b2+b3b4+...   bn>0 satisfies, (i) bn+1bn   for all n and (ii) limnbn=0 , then the series is convergent”

Calculation:

Let an=xnn2 .

Then an+1=xn+1(n+1)2 .

Obtain |an+1an| to apply the Ratio test.

|an+1an|=|xn+1(n+1)2xnn2|

Take limn on both sides,

limn|an+1an|=limn|xn+1(n+1)2xnn2|=limn|n2xnxn+1(n+1)2|=limn|n2xn2+2n+1|=limn|x1+2n+1n2|

Apply the limit and simplify the terms as shown below.

limn|x1+2n+1n2|=|x1+2+12|=|x1+0+0|=|x|

The series converges when the limit is less than

The series n=1xnn2 converges as |x|<1 .

|x|<11<x<1

Consider in case of x=1 , then the series n=1xnn2 expressed as follows:

n=1xnn2=n=1(1)nn2

By using alternating series test, Let bn=1n2

n2<n2+11n2+1<1n2bn+1<bn

That is, bn is decreasing.

Obtain the limit bn .

limnbn=limn1n2=1=0

Thus, the limit limnbn=0 .

Therefore, by the result (4) the series is n=1xnn2 converges when x=1 .

Let x=1 be the end point of the interval. Then, the series expressed as follows:

n=1xnn2=n=1(1)nn2=n=11n2

Hence, by the result (2) n=11n2 is converges. since, which is a p-series with p=2 .

Therefore, n=1xnn2 is converges when x=1 .

Hence, the interval of the convergence is I=[1,1] .

Differentiate f(x)=n=1xnn2 as shown below,

f(x)=n=1nxn1n2=n=1xn1n

Apply the ratio test,

Let an=xn1n .

Then, an+1=x(n+1)1(n+1) .

Obtain |an+1an| to apply the Ratio test.

|an+1an|=|x(n+1)1(n+1)xn1n|

Take limn on both sides,

limn|an+1an|=limn|x(n+1)1(n+1)xn1n|=limn|xn(n+1)nxn1|=limn|n(n+1)1x1|=limn|x|(nn+1)

Simplify further,

limn|an+1an|=limn|x|(nn+1)=limn|x|(11+1n)

Apply the limit and simplify the terms as shown below.

limn|x|(11+1n)=|x|(11+1)=|x|(11+0)=|x|

The series converges when the limit is less than 1.

The series n=1xnn2 converges as |x|<1

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