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Find the radius and the interval of convergence for the following power series. Justify your answer!

$\sum _{ k=1 }^{ \infty }{ \frac { (-1)^ k+1(x-1)^ k }{ k3^ k } } $

My attempt

First by using ratio test, I got $-2<x<4$

If x=-2, then $\sum _{ k=1 }^{ \infty }{ \frac { (-1)^ k+1(-3)^ k }{ k3^ k } } $ = $\sum _{ k=1 }^{ \infty }{ \frac { (-1)^ 2k }{ k } } $

I am stuck here... can anyone show how to do after substitution and find the interval of convergence.

tien lee
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1 Answers1

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For $x = -2$, we get $$ \sum_{k = 1}^\infty \frac{(-1)^k + (-3)^k}{k3^k} = \sum_{k = 1}^\infty \frac{(-1)^k(1 + 3^k)}{k3^k} = \sum_{k = 1}^\infty \frac{(-1)^k}{k3^k} + \sum_{k = 1}^\infty\frac{(-1)^k}{k}, $$ which converges by the alternating series test. For the $x = 4$ case, you'll get a non-alternating series, and won't be able to conclude this.

Chris
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