Prove $\sum_{n=1}^\infty \frac{nz^n}{1-z^n}=\sum_{n=1}^\infty \frac{z^n}{(1-z^n)^2}$

Question

Prove that $$\sum_{n=1}^\infty \frac{nz^n}{1-z^n}=\sum_{n=1}^\infty \frac{z^n}{(1-z^n)^2}$$ and give the region where is an holomorphic function.

I already know that are holomorphic functions in the open disk $\{z\in \mathbb{C}:|z|<1\}$ but I don't know how to prove the equality.

Answer 1

\begin{align} \sum_{n\ge1}\frac{nz^n}{1-z^n}&=\sum_{n\ge1}n\sum_{k\ge1}z^{nk}\\ &=\sum_{k\ge1}\sum_{n\ge1}nz^{nk}\\ &=\sum_{k\ge1}\frac{z^k}{(1-z^k)^2}, \end{align} where you can exchange the order of sums because of absolute convergence.