Convergence of $\sum_{n=0}^{\infty}\frac{z^n}{1+z^{2n}}$

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 $|z|\ne 1$.

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

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.