   Chapter 11.8, Problem 15E

Chapter
Section
Textbook Problem

# Find the radius of convergence and interval of convergence of the series.15. ∑ n = 0 ∞ ( x − 2 ) n n 2 + 1

To determine

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

Explanation

Given:

The series n=1(x2)nn2+1 .

Result used:

(1) Ratio test: If limn|an+1an|=L<1 , then the series n=1an is absolutely convergent.

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

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

(4)“Suppose that an and bn are series with positive terms

(i)if bn is convergent and anbn for all n, then an is also convergent.

(ii) if bn is divergent and anbn for all n, then an is also divergent.”

Calculation:

Let an=(x2)nn2+1 .

Then, an+1=(x2)n+1(n+1)2+1 .

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

|an+1an|=|(x2)n+1(n+1)2+1(x2)nn2+1|

Take limn on both sides,

limn|an+1an|=limn|(x2)n+1(n+1)2+1(x2)nn2+1|=limn[(x2)n+1(n+1)2+1.n2+1(x2)n]=limn[|x2|.n2+1(n+1)2+1]

Simplify the terms of the expression,

limn|an+1an|=limn[|x2|.n2+1n2(n+1)2+1n2]=limn[|x2|.n2n2+1n2(n+1)2n2+1n2]=limn[|x2|.1+1n2(n+1)2n2+1n2]

Apply the limit and simplify the terms as shown below.

limn|x2|1+1n2(1+1n)2+1n2=[|x2|1+12(1+1)2+12]=[|x2|1+0(1+0)2+0]=[|x2|1(1)2]=|x2|

The series n=1(x2)nn2+1 converges as |x2|<1

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