For which values of $x$ does $\sum_{n=1}^{\infty}\frac{ x^n}{x^{2n} -1}$ converge? Of course it does not converge for $x = \pm 1$. I tried to use the limit form of the comparison test, using a divisor "b" series of $x^n$, and $\frac{1}{x^{2n} -1}$ goes to $0$ as $n$ goes to $\infty$ for $|x| > 1$, BUT my test series converges only for $|x| < 1$ so I am likely using this incorrectly. for $|x| < 1$ I don't know how to proceed. 
 A: If we set $f_n(x)=\frac{x^n}{x^{2n}-1}$ we have that $f_n(x)=- f_n\left(\frac{1}{x}\right)$, so it is enough to prove convergence over $x\in(-1,1)$ to have convergence over $\mathbb{R}\setminus\{-1,+1\}$. But if $|x|<1$ we have:
$$ \frac{x^n}{1-x^{2n}} = x^{n}+x^{3n}+x^{5n}+\ldots $$
so:
$$ \sum_{n\geq 1}\frac{x^n}{1-x^{2n}} = \sum_{n\geq 1} \bar{\sigma}(n)\,x^n $$
where $\bar{\sigma}(n)$ is the number of odd divisors of $n$. But obviously $1\leq\bar{\sigma}(n)\leq n$, hence the radius of convergence of the previous power series is one and we are done.
A: Observe that $\;x^{2n}-1=(x^n-1)(x^n+1)\;$ , so in fact
$$\frac{x^n}{x^{2n}-1}=\frac{x^n}{\left(x^n-1\right)\left(x^n+1\right)}=\frac12\left(\frac1{x^n-1}+\frac1{x^n+1}\right)$$
Try now to attack it from here. 
Added by request: for $\;|x|<1\;$ we have the limit comparison theorem with $\;\cfrac1{|x|^n}\;$ :
$$\frac{\cfrac1{|x^n\pm1|}}{\cfrac1{|x^n|}}=\frac{|x|^n}{|x^n\pm1|}\xrightarrow[n\to\infty]{}0$$
and since the geometric series $\;\sum\limits_{n=0}^\infty\cfrac1{x^n}\;$ converges absolutely for $\;|x|<1\;$, so do both our series. 
$$\sum_{n=0}^\infty\frac1{x^n\pm1}$$
Observe we used above the strong one-direction limit comparison test for non-negative series:
$$\text{If}\;\;\lim_{n\to\infty}\frac{a_n}{b_n}=0\;\;\text{ and}\;\;\sum_{n=0}^\infty b_n\;\;\text{converges, then so does}\;\;\sum_{n=0}^\infty a_n$$
A: Hint 1: For $\lvert x\rvert<1$, use ratio test and see what you get.
Hint 2: For $\lvert x\rvert>1$, use ratio test and see what you get.
