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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 $|z|<1$?

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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:

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

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    Thanks for your response. Could you expand more on how to use the hints?2011-11-19
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    @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.2011-11-19
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    That was really hard to enter using the mobile interface.2011-11-19
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    @Didier: Thanks2011-11-19
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    @robjohn: I'm sorry for putting you through the trouble..:) Thanks very much for your help.2011-11-19
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    @Chris: You're welcome. It was my choice to try to answer while away from home. I knew it was not going to be as easy as with a mouse and full keyboard.2011-11-19