For what values does $\sum_0^\infty\frac{z^n}{1+z^{2n}}$ converge? Was playing around the series $\sum_0^\infty\frac{z^n}{1+z^{2n}}$, where $z$ is complex, trying to figure out where it converges.
Assuming $|z|>1$
$$
\frac{|z^n|}{|1+z^{2n}|}>\frac{|z^n|}{1+|z^{2n}|}>\frac{|z^n|}{2|z|^{2n}}=\frac{1}{2|z|^{n}}
$$
but that didn't get me far, since the series whose terms are the last number above is a convergent series. 
What is a better way to approach this series to determine where it converges? Thanks.
 A: Given $z \in \mathbb{C}$ with $|z| > 1$, choose any $r > 1$. We have
$$|1 + z^{2n}| \geq \left|1 - |z|^{2n}\right| > \frac{|z|}{r}^{2n}$$
for large enough $n$. The first inequality comes from $|a + b| \geq ||a| - |b||$, which is always true, and the second one comes from the limit
$$\lim_{n \rightarrow \infty} \frac{\left|1 - |z|^{2n}\right|}{|z|^{2n}} = 1,$$
and the choice of $r$, which makes $1/r$ less than $1$. Now, applying the root test as azarel suggested, we get
$$\limsup_n \sqrt[n]{\left|\frac{z^n}{1 + z^{2n}}\right|} < \limsup_n \frac{|z|}{\frac{|z|^2}{\sqrt[n]{r}}} = \frac{1}{|z|} < 1.$$
Therefore the series converges for $|z| > 1$.
As Greg Martin noted, this series may be written as
$$\sum_{n = 0}^{\infty} \frac{1}{z^n + z^{-n}}$$
which shows that its behavior must be the same for $|z| > 1$ or $|z| < 1$, so that we get convergence on $\mathbb{C}\backslash\mathbb{S}^1$. The behavior for $|z| = 1$ seems more complicated, though, so I'll leave to someone more experienced to answer. Hope this helps!
A: The cases $|z|\ne1$ having been disposed of, note that when $|z|=1$, the denominator is bounded above (in absolute value) by 2, while the numerator is (in absolute value) 1, so the quotient is bounded below (in absolute value) by $1/2$. In particular, the terms are not going to zero, so the series diverges. 
