   Chapter 11.8, Problem 6E

Chapter
Section
Textbook Problem

# Find the radius of convergence and interval of convergence of the series.6. ∑ n = 1 ∞ ( − 1 ) n x n n 2

To determine

To find: The radius of convergence and interval of convergence of the series.

Explanation

Given:

The series is n=1(1)nxnn2 .

Ratio test:

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

Alternating series test:

“If the alternating series n=1(1)n1bn=b1b2+b3b4+b5b6+.... satisfies the conditions, bn+1bn and limxbn=0 , then the series is convergent.”

Result used:

The p-series is convergent n=11np is convergent if p>1 and divergent if p1 .

Calculation:

Let an=(1)nxnn2 .

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

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

|an+1an|=|(1)n+1xn+1(n+1)2.n2(1)nxn|

Take limn on both sides,

limn|an+1an|=limn|(1)n+1xn+1(n+1)2.n2(1)nxn|=limn|(1)xn2(n+1)2|=limn[|x|n2(n+1)2]

Divide both the numerator and denominator by n2 ,

limn|an+1an|=limn[|x|n2n2n2n2+2nn2+1n2]=limn[|x|1+2n+1n2]

Apply the limit and simplify the terms as shown below.

limn[|x|1+2n+1n2]=[|x|1+2+12]=[|x|1+0+0]=|x|

The series n=1(1)nxnn2 converges as |x|<1

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