Does $\sum_{n=2}^{\infty} \frac{n}{x^n-1}$ converge?

I'm unsure for what $$x$$ the following series converges/diverges:

$$\sum_{n=2}^{\infty} \frac{n}{x^n-1}$$

WolframAlpha is neither able to find an answer (with standard computing time, does anyone have WolframAlpha Pro?). I'm not that experienced with determining whether serieses converge or not but I tried to apply the ratio test:

\begin{align} L &= \lim _{n\to \infty}|{\frac {a_{n+1}}{a_{n}}}| \\\\ &= \lim _{n\to \infty}|{\frac {(n+1)x^n-n-1}{nx^{n+1}-n}}| \end{align}

...but I don't know how to continue from here.

So does this series converge or diverge (and for what $$x$$)? Any ideas?

• What happens for different values of $x$ ? Are there any values you can immediately discount because of the limit of the series? – Ninad Munshi Apr 16 at 11:33

If $$|x| <1$$ then $$\frac n {x^{n}-1}$$ does not tend to $$0$$. Hence the series does not converge in this case. If $$|x|>1$$ then $$|\frac n {x^{n}-1} |\leq \frac {2n} {|x|^{n}}$$ for $$n$$ sufficiently large and $$\sum \frac {2n} {|x|^{n}} <\infty$$ by ratio test. I leave the case $$x=1$$ and $$x=-1$$ to you.