An exercise of complex series

Question

Prove that $\sum_{n=1}^\infty \cfrac{nz^n}{1-z^n}=\sum_{n=1}^{\infty}\cfrac{z^n}{(1-z^n)^2}$ for $|z|<1$.

The hint is to develop in a double series and reverse the order of summation. But I tried a lot and don't know which part to expand.

Answer 1

We shall use the identity \begin{align} \frac{z}{(1-z)^2} = \sum^\infty_{n=1} nz^{n}. \end{align} Observe we have \begin{align} \sum^\infty_{n=1} \frac{z^n}{(1-z^n)^2} = \sum^\infty_{n=1}\sum^\infty_{k=1}kz^{nk} = \sum^\infty_{k=1}k\sum^\infty_{n=1}z^{nk} = \sum^\infty_{k=1}k\frac{z^{k}}{1-z^k}. \end{align}