   Chapter 11.8, Problem 5E

Chapter
Section
Textbook Problem

# Find the radius of convergence and interval of convergence of the series.5. ∑ n = 1 ∞ x n 2 n − 1

To determine

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

Explanation

Given:

The series is n=1xn2n1 .

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 .

Limit comparison test:

“Suppose that an and bn are the series with positive terms, if limnanbn=c , where c is a finite number and c>0 , then either both series converge or both diverge.”

Calculation:

Let an=xn2n1

Then, an+1=xn+12n+1 .

Obtain |an+1an| .

|an+1an|=|xn+12(n+1)1.2n1xn|

Take limn on both sides,

limn|an+1an|=limn|xn+12(n+1)1.2n1xn|=limn|xn.x2n+1.2n1xn|=limn|x(2n1)2n+1.1n1n|=limn|x(21n)2+1n|

Apply the Ratio test and simplify the terms as shown below.

limn[(21n)|x|2+1n]=[212+1.|x|]=|202+0.|x||=|x|

Thus, the series n=1xn2n1 converges as |x|<1 .

Therefore, the radius of convergence is R=1

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