# 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 $|z|<1$?

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

## 1 Answer

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? – Chris Nov 19 '11 at 17:05
@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
That was really hard to enter using the mobile interface. – robjohn Nov 19 '11 at 17:59
@Didier: Thanks – Chris Nov 19 '11 at 19:30
@robjohn: I'm sorry for putting you through the trouble..:) Thanks very much for your help. – Chris Nov 19 '11 at 19:31