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 $|z|<1$?

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:
$$
\begin{align}
\sum_{n=1}^\infty\frac{z^n}{(1-z^n)^2}
&=\sum_{n=1}^\infty\sum_{k=0}^\infty(k+1)z^{kn+n}\\
&=\sum_{n=1}^\infty\sum_{k=1}^\infty kz^{kn}\\
&=\sum_{k=1}^\infty\sum_{n=1}^\infty kz^{kn}\\
&=\sum_{k=1}^\infty\sum_{n=0}^\infty kz^{kn+k}\\
&=\sum_{k=1}^\infty\frac{kz^k}{1-z^k}
\end{align}
$$
