Convergence of $\sum_{n=0}^{\infty}\frac{z^n}{1+z^{2n}}$ 
For what complex values of $z$ is $$\sum_{n=0}^{\infty}\dfrac{z^n}{1+z^{2n}}$$ convergent?

I would like to write the sum as a power series, because with a power series we can determine the radius of convergence. But in this case it seems untidy. We have $\dfrac{1}{1+z^{2n}}=1-z^{2n}+z^{4n}-\cdots$, so that the original sum is $$\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}(-1)^kz^{n+2nk}$$ and I don't think that's very useful...
 A: Hint: If $|z|\lt 1$, it should be a straightforward comparison, after a while the norm of the $n$-th term is $\lt \frac{1}{2}|z|^n$.  
If $|z|=1$, the terms do not approach $0$, so we cannot have convergence. 
And if $|z|\gt 1$ it should be a straightforward comparison: divide top and bottom by $z^n$. 
The conclusion will be that we have convergence if $|z|\ne 1$.
A: Hint: When $|z|<1$ we have $$\left|\frac{z^n}{1+z^{2n}}\right| \le \frac{|z|^n}{1-|z|^{2}},$$
when $|z|>1$ we have
$$\left|\frac{z^n}{1+z^{2n}}\right| \sim {|z|^{-n}}$$
and when $|z|=1$ we have
$$\left|\frac{z^n}{1+z^{2n}}\right| \not\to 0.$$
A: This looks like a good time for the ratio test:
Defining $a_n=\frac{z^n}{1+z^{2n}}$, we have
$$
\begin{align}
\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| &=
\lim_{n\to\infty} \frac{z^{n+1}}{1+z^{2(n+1)}} \frac{1+z^{2n}}{z^n}\\
&= \lim_{n\to\infty} z\frac{1+z^{2n}}{1+z^{2(n+1)}}\\
&= \lim_{n\to\infty} \frac{z+z^{2n+1}}{1+z^{2(n+1)}}\\
&= \lim_{n\to\infty} \frac{z^{-2n-1}+z^{-1}}{z^{-2n-2}+1}
\end{align}
$$
When $|z|\neq 1$, the above approaches zero, implying convergence. Otherwise, the test is inconclusive.
In fact, we can conclude that the terms of the sum do not converge to zero when $|z|=1$ since by the triangle inequality, we have
$$
\begin{align}
|a_n| &= \left| \frac{z^n}{1+z^{2n}}\right|\\
&=\frac{|z|^n}{|1+z^{2n}|}\\
&\geq \frac{|z|^n}{|1|+|z|^{2n}} 
= \frac{1^n}{1+1^{2n}}=\frac12
\end{align}
$$
Which means that the sum must diverge for all such values.
A: If $|z|<1$ then
$$\left|\frac{z^n}{1+z^{2n}}\right|\leq \frac{|z|^n}{1-|z|^{2n}}\sim_\infty |z|^n$$
and the geometric series $\sum_n |z|^n$ is convergent so the given series is absolutely convergent.
If $|z|>1$ then
$$\left|\frac{z^n}{1+z^{2n}}\right|\leq \frac{|z|^n}{|z|^{2n}-1}\sim_\infty |z|^{-n}$$
and the geometric series $\sum_n |z|^{-n}$ is convergent so the given series is absolutely convergent.
If $|z|=1$ then the series is divergent since the general term doesn't converge to $0$.
