$\sum_{n=1}^{\infty} nz^n(1-z^n)^{-1} = \sum_{n=1}^{\infty} z^n(1-z^n)^{-2}$ for $|z|<1$ I want to prove the equality $\sum_{n=1}^{\infty} nz^n(1-z^n)^{-1} = \sum_{n=1}^{\infty} z^n(1-z^n)^{-2}$ for $|z|<1$. This is an exercise in Ahlfors's complex analysis. There is a hint given by:
Develop in a double series and reverse the order of summation.
But I have no idea. How do I have to proceed?

$$
\eqalign{
  & \sum\limits_{1\, \le \,n} {{{n\,z^{\,n} } \over {1 - \,z^{\,n} }}}  = \sum\limits_{0\, \le \,\left( {1\, \le } \right)\,n} {n\,z^{\,n} \sum\limits_{0\, \le \,k} {z^{\,n\,k} } }
  = \sum\limits_{0\, \le \,k} {\sum\limits_{0\, \le \,\left( {1\, \le } \right)\,n} {n\,z^{\,n\left( {k + 1} \right)} } }  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {z^{\,\left( {k + 1} \right)} \sum\limits_{0\, \le \,\left( {1\, \le } \right)\,n} {n\,\left( {z^{\,\left( {k + 1} \right)} } \right)^{\,n - 1} } }
  = \sum\limits_{0\, \le \,k} {{{z^{\,\left( {k + 1} \right)} } \over {\left( {1 - z^{\,\left( {k + 1} \right)} } \right)^2 }}}   \cr 
  &  = \sum\limits_{1\, \le \,k} {{{z^{\,k} } \over {\left( {1 - z^{\,k} } \right)^2 }}}  \cr} 
$$
