The following identity arose while I was working on a recent [MO question](http://mathoverflow.net/questions/108927/a-question-about-a-formal-power-series-manipulation):


$-\sum\_{n=1}^{\infty}\frac{1}{n}\frac{(-x)^n}{1-x^n}=\sum\_{n=1}^{\infty}\frac{1}{n}\frac{x^n}{1-x^{2n}}.$


I have no doubt that the identity is true, but I am not able to prove it. Can anyone help?


It is easy to prove by Taylor expansion that the left-hand-side of the identity can equivalently be written as $\sum\_{n=1}^{\infty}\ln(1+x^n)$, which is the logarithm of the [q-Pochhammer symbol](http://en.wikipedia.org/wiki/Q-Pochhammer_symbol) $(-x,x)\_{\infty}$, so an alternative way to pose my question is to ask for a proof of the series expansion


$\ln(-x,x)\_{\infty}=\sum\_{n=1}^{\infty}\frac{1}{n}\frac{x^{n}}{1-x^{2n}}.$



series expansion of the q-Pochhammer symbol
First notice that 
$$\sum \_{n=1} ^{\infty} \frac{x^n}{n(1-x^{2n})} = \sum \_{r=0} ^{\infty} \sum \_{m=1} ^{\infty}\left(\frac{1}{2^r}\sum \_{k|2m-1} \frac{1}{k}\right)x^{2^r(2m-1)}.$$
And similarly
$$-\sum \_{n=1}^{\infty}\frac{(-x)^n}{n(1-x^n)} = \sum \_{s=1}^{\infty} \left(\sum \_{k|s}\frac{(-1)^{k+1}}{k}\right)x^s.$$
So we need to show that the respective coefficients match, i.e.: 
$$\frac{1}{2^r}\sum \_{k|2m-1} \frac{1}{k}=\sum \_{k|s}\frac{(-1)^{k+1}}{k},$$
for $s=2^r(2m-1)$. But this is a simple corollary of $\frac{1}{2^r}=1-(\frac{1}{2}+\cdots+\frac{1}{2^r})$.


