   Chapter 11.8, Problem 11E

Chapter
Section
Textbook Problem

# Find the radius of convergence and interval of convergence of the series.11. ∑ n = 1 ∞ ( − 1 ) n 4 n n x n

To determine

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

Explanation

Given:

The series n=1(1)n4nnxn .

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)n4nnxn .

Then, an=(1)n+14n+1n+1xn+1 .

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

|an+1an|=|(1)n+14n+1n+1xn+1(1)n4nnxn|

Take limn on both sides,

limn|an+1an|=limn|(1)n+14n+1n+1xn+1(1)n4nnxn|=limn|(1)n+14n+1n+1xn+1.n(1)n4nxn|=limn[4n+1|x|n+1n+1.n4n|x|n]=limn4|x|.nn+1

Divide both the numerator and denominator by n ,

limn|an+1an|=limn4|x|.nnn+1n=limn4|x|11+1n

Apply the limit and simplify the terms as shown below.

limn4|x|11+1n=4|x|11+1=4|x|1=4|x|

The series n=1(1)n4nnxn converges as |x|<1

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