I have a given alternating power series: $$\sum\limits_{n=2}^{\infty}(-1)^{n}\frac{1}{n2^n}\cdot x^n$$ and I need to find the Radius of Convergence.

I tried myself with this task and I found an answer, but I am not sure, if I am right.

I can easily see, the series is alternating. So I thought about using the alternating series test (Leibniz Criterion) to see, if this series converge.

I calculated: $$\lim_{n \to \infty} \frac{1}{n2^n} = 0$$ and this is obviously right: $$\frac{1}{n2^n}\geq \frac{1}{(n+1)2^{(n+1)}}$$

So I think, I can be sure, the series converges, but what about the Radius of Convergence? I thought about using this, to determine the Radius, $r$:

$$r =\ \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}}\right|.$$

Applied on my given series:

$$\left| \frac{a_{n}}{a_{n+1}}\right| = \frac{(n+1)2^{n+1}}{n2^n}=2\frac{n+1}{n} \longrightarrow 2$$

So, that's why I think, the Radius of Converges must be $\lvert x \rvert \leq 2$

Am I right?

PS: made some edits to improve the latex

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    $\begingroup$ Yes, the radius of convergence is $2$. However, convergence at the boundary, i.e. for $x=2$ and $x=-2$ must be checked separately. (Turns out the series converges to $\frac x2-\ln(1+\frac x2)$) $\endgroup$ – Hagen von Eitzen May 6 '13 at 16:18
  • $\begingroup$ oh okay, i have completly forgotten the boundaries. thanks $\endgroup$ – Toralf Westström May 6 '13 at 16:35
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    $\begingroup$ Oh, excuse me: how did you find out the that the series converges towards $\frac{x}{2}-ln(1+ \frac{2}{x})$ ? Your Method? $\endgroup$ – Toralf Westström May 6 '13 at 17:23

Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly.

You are guaranteed that your series converges for $|x| \lt 2$: i.e. $2$ is the radius of convergence. What happens at $x = 2$ and at $x = -2$ then needs to be checked, if we want to know what happens at at the bounds of the radius of convergence.

  • $\begingroup$ Nice advice! +1 $\endgroup$ – Amzoti May 7 '13 at 0:29

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