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...

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
$$\left|\frac{z^n}{1+z^{2n}}\right| \not\to 0.$$
