Analyze convergence of the series $\sum_{n=1}^\infty \frac{1}{r^n - \frac{1}{r^n}}$ for $r \in (0, \infty) \setminus \{1\}$ I would like to have a little help determining for which $r \in (0, \infty) \setminus \{1\}$ the following series converges:
$$\displaystyle \sum_{n=1}^\infty \frac{1}{r^n - \frac{1}{r^n}}.$$
I'm having trouble with the fact that there are two terms in the denominator which appear to be "fighting" against each other. For instance, when $r > 1$, the series above is painfully close the geometric series $\displaystyle\sum_{n=1}^\infty\frac{1}{r^n}$. But unfortunately, the $-\frac{1}{r^n}$ in the denominator makes each term $\frac{1}{r^n - \frac{1}{r^n}}$  somewhat larger than it's counterpart in the geometric series.
My ultimate goal is to have the real series $\displaystyle \sum_{n=1}^\infty \frac{1}{r^n - \frac{1}{r^n}}$ help me analyze the convergence of the complex series
$$\displaystyle \sum_{n=0}^\infty\frac{z^n}{1 + z^{2n}} =\sum_{n=0}^\infty\frac{1}{z^n + \frac{1}{z^{n}}}, $$
for $z \neq 0$ and $|z| \neq 1$. The connection I see between the real and complex series comes from the triangle inequality:
$$\displaystyle \frac{1}{|z^n + \frac{1}{z^{n}}|} \le \frac{1}{||z|^n - \frac{1}{|z|^n}|},$$
for $n \ge 1$. Suggestions or solutions are greatly appreciated.

If $r>1$ first, then your general term is nonnegative and 
$$
\frac{1}{r^n(1-1/r^{2n})}\sim \frac{1}{r^n}=(1/r)^n
$$
so the series converges since $0\leq 1/r<1$ (geometric series on the right).
If $r<1$ then, then the general term is
$$
\frac{r^n}{r^{2n}-1}\sim-r^n
$$
so the series converges again since $0\leq r<1$ (geometric series on the right again).
Note: this means in particular, thanks to your work, that the complex series you are ultimately interested in converges absolutely for every $|z|\neq 1$.
