# Lambert series expansion identity

I have a question which goes like this:

How can I show that $$\sum_{n=1}^{\infty} \frac{z^n}{\left(1-z^n\right)^2} =\sum_{n=1}^\infty \frac{nz^n}{1-z^n}$$ for $|z|<1$?

-
Thanks for the info. –  Chris Nov 19 '11 at 17:04

Hint: Try using the expansions $$\frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+\dots$$ and $$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+5x^4+\dots$$ Expansion:
\begin{align} \sum_{n=1}^\infty\frac{z^n}{(1-z^n)^2} &=\sum_{n=1}^\infty\sum_{k=0}^\infty(k+1)z^{kn+n}\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty kz^{kn}\\ &=\sum_{k=1}^\infty\sum_{n=1}^\infty kz^{kn}\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty kz^{kn+k}\\ &=\sum_{k=1}^\infty\frac{kz^k}{1-z^k} \end{align}
@Chris: Try to plug $x=z^n$ in the $n$th terms of the LHS and of the RHS of the relation you want to prove. –  Did Nov 19 '11 at 17:44