For which values of $x$ does the following series converge?

Question

For which values of $x$ does the following series $$\sum \frac{x^n}{x^{2n}-1}$$ converge?

The author in the solution says that for $-1<x<1$ the series converges because $\sum x^n$ is convergent and $x^{2n}\to 0$ as $n\to \infty.$ I am not sure that I understand this reasoning. Is he using the fact that if $\sum a_n$ is a convergent series and the $b_n$ is a bounded sequence then $\sum a_n b_n$ is convergent provided $a_n\geq0.$ This may work for $0\leq x< 1,$ but I how can he use this fact when the terms of $x^n$ may be negative?

Answer 1

$\sum{x^n}$ is absolutely convergent. This should resolve your difficulties, I think.

Answer 2

In fact the series converges for $x\ne1,-1$.

If $-1<x<1$ we've seen that the series converges.

Now suppose $|x|>1$. There exists $N$ so that $$x^{2n}-1\ge\frac12 x^{2n},\quad(n>N).$$Hence for $n>N$ we have$$\left|\frac{x^n}{x^{2n}-1}\right|\le 2|x|^{-n}.$$