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...

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 $0$.
