series expansion of the q-Pochhammer symbol - MathOverflow

series expansion of the q-Pochhammer symbol

2012-11-06T14:36:40Z

The following identity arose while I was working on a recent MO question:

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

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 $(-x,x)_{\infty}$, so an alternative way to pose my question is to ask for a proof of the series expansion

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

Answer by Gjergji Zaimi for series expansion of the q-Pochhammer symbol

2012-11-06T15:10:13Z

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 $\frac{1}{2^r}=1-(\frac{1}{2}+\cdots+\frac{1}{2^r})$.

Answer by Garth Payne for series expansion of the q-Pochhammer symbol

2012-11-06T19:06:09Z

I would make a mere comment since Gjergji has already answered, but I am not allowed to make comments.

... so an alternative way to pose my question is to ask for a proof of the series expansion $\ln(-x,x)_{\infty}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^{n}}{1-x^{2n}}$ .

This is a corollary of Euler's theorem that the number of partitions of $n$ into distinct parts is equal to the number of partitions of $n$ into odd parts. In terms of generating functions, Euler's theorem is just $(-x,x)_{\infty}=\frac{1}{(x,x^2)_\infty}$ , which can be easily proved by replacing the term $(1+x^i)$ in $(-x,x)_\infty$ by $\frac{1-x^{2i}}{1-x^i}$ and cancelling all the terms in the numerator against the corresponding terms in the denominator. By Euler's theorem, $\ln \left( (-x,x)_{\infty}\right) =\ln\left( \frac{1}{(x,x^2)_\infty}\right) =\sum_{i=1}^\infty \ln \left( \frac{1}{1-x^{2i-1}} \right) =\sum_{i=1}^\infty \sum_{n=1}^\infty \frac 1 n x^{n (2i-1)}$

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