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 $|z|<1$?
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} $