Prove that $\sum_0^\infty \frac{z^n}{1-z^{2n}}$ converges for all $\left|z\right|<1$ and $\left|z\right|>1$. Question:

Prove that $\sum_0^\infty \frac{z^n}{1-z^{2n}}$ converges for all $\left|z\right|<1$ and $\left|z\right|>1$.

Attempt:
For the case $\left|z\right|>1$, I was hoping to use the comparison test somehow against $\frac{1}{\left|z\right|^n}$ but I only got this far:
\begin{align*}
\frac{\left|z\right|}{\left|1-z^{2n}\right|} = \frac{\left|z\right|}{(1+\left|z\right|^n)(1-\left|z\right|^n)}
< \frac{\left|z\right|^n + 1}{(1+\left|z\right|^n)(1-\left|z\right|^n)}
< \frac{2}{-\left|z\right|^n}
\end{align*}
 A: You have claimed for (for $|z|>1$) that $$\frac{\left|z\right|}{\left|1-z^{2n}\right|} = \frac{\left|z\right|}{(1+\left|z\right|^n)(1-\left|z\right|^n)}$$ however this is not true, because the quantity on the left is strictly positive while the quantity on the right is strictly negative (because $1-|z|^n$ is negative for all $|z|>1$ and $n>0$). To fix this, I believe we should have  $$\frac{\left|z\right|}{\left|1-z^{2n}\right|} = \frac{\left|z\right|}{|1+\left|z\right|^n||1-\left|z\right|^n|}$$
where now we observe that $$|1-|z|^n| = |z|^n-1 \quad \text {and} \quad |1+|z|^n|=1+|z|^n$$
so we end up with instead $$\frac{\left|z\right|}{\left|1-z^{2n}\right|} = \frac{\left|z\right|}{(1+|z|^n)(|z|^n-1)} $$
Now let's use the comparison test. $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty}\frac{\frac{\left|z\right|}{\left|1-z^{2(n+1)}\right|}}{\frac{\left|z\right|}{\left|1-z^{2n}\right|}}\\  =\lim_{n \to \infty} \frac{\left|1-z^{2n}\right|}{\left|1-z^{2n+2}\right|} \\ = \lim_{n \to \infty}\frac{(1+|z|^n)(|z|^n-1)}{(1+|z|^{n+1})(|z|^{n+1}-1)} \\ < \lim_{n \to \infty}\frac{(1+|z|^{n})(|z|^{n+1}-1)}{(1+|z|^{n+1})(|z|^{n+1}-1)} \\ =  \lim_{n \to \infty}\frac{1+|z|^{n}}{1+|z|^{n+1}} \\ < \lim_{n \to \infty}\frac{1+|z|^{n}}{|z|^{n+1}} \\= \lim_{n \to \infty}\frac{1}{|z|}+ \frac{1}{|z|^{n+1}} \\ = \frac{1}{|z|} \\ <1$$
So the series converges for $|z|>1$.
For the case of $|z|<1$, your equality $$\frac{\left|z\right|}{\left|1-z^{2n}\right|} = \frac{\left|z\right|}{(1+\left|z\right|^n)(1-\left|z\right|^n)}$$ does hold because $1-|z|^n$ is always positive. If you play around with the inequalities some more you should be able to figure out how your series behaves for $|z|<1$. Someone gave the hint to use $|z|=\frac{1}{|w|}$ which is a good one. I will leave the rest to you!
