Is $\sum_{n=1}^{\infty}\frac{z^{2n}}{1-z^{2n}}$ convergent?

Question

Is $$\sum_{n=1}^{\infty}\frac{z^{2n}}{1-z^{2n}}, \hspace{1cm} |z| < 1$$ convergent? I think it is convergent and tried expanding the denominator but did not get any result. Thanks for your help.

Answer 1

Suppose $|z_0|=r<1$, then $$\left|\frac{z_0^{2n}}{1-z_0^{2n}}\right|\leq\frac{|z_0|^{2n}}{1-|z_0|^{2n}} <2|z_0|^{2n}=r^{2n},$$ when $n$ is large enough! (due to $\lim_{n\to \infty}|z_0|^{2n}=0$, then there exists $N$ such that when $n>N$, we have $|z_0|^{2n}<\frac{1}{2}$.). So the series $$\sum\left|\frac{z_0^{2n}}{1-z_0^{2n}}\right|$$ is convergent, $$\sum\frac{z_0^{2n}}{1-z_0^{2n}}$$ is absolutely convergent, hence convergent.

Answer 2

Write $a_n=z^{2n}/(1-z^{2n})$. Then $$\frac{a_{n+1}}{a_n}=z^2\frac{1-z^{2n}}{1-z^{2n+2}}$$ which converges to $z^2$ whenever $|z|<1$.

Answer 3

If $|z|<1$ then $|z^{2n}|<1/2$ for all large enough $n$. Therefore $ |1-z^{2n}| > 1/2$ for all large enough $n$ so the sum behaves like $z^{2n}$ which converges for $|z|<1$.