   Chapter 11.8, Problem 9E

Chapter
Section
Textbook Problem

# Find the radius of convergence and interval of convergence of the series.9. ∑ n = 1 ∞ x n n 4 4 a

To determine

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

Explanation

Given:

The series n=1xnn44n .

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=xnn44n .

Then, an+1=xn+1(n+1)44n+1 .

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

|an+1an|=|xn+1(n+1)44n+1xnn44n|

Take limn on both sides,

limn|an+1an|=limn|xn+1(n+1)44n+1xnn44n|=limn|xn+1(n+1)44n+1.n44nxn|=limn|x|n+1(n+1)44n+1.n44n|x|n=limn|x|4.(nn+1)4

Divide both the numerator and denominator by n ,

limn|an+1an|=limn|x|4.(nn1n+nn)=limn(11n+1)4

Apply the limit and simplify the terms as shown below

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