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


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


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2 Answers 2

up vote 6 down vote accepted

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

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answered within 33 minutes --- MO rules! – Carlo Beenakker Nov 6 '12 at 15:22
@Gjergji Zaimi Thanks for the help. I could not derive any of the 2 equalities that you stated in the beginning. Can you kindly give some proof or reference for them? Especially in the second one I don't see any occurrence of the summing variable k in the summand. Can you kindly add some more lines of explanation! – Anirbit Nov 19 '12 at 23:56
Anirbit, thanks for catching the typo. I fixed it. – Gjergji Zaimi Nov 20 '12 at 1:51
I may try to add some more explanations a bit later, but if you expand the fractions $\frac{1}{1-x^n}=1+x^n+x^{2n}+\cdots$, it shouldn't be very hard to see the equations that I wrote. – Gjergji Zaimi Nov 20 '12 at 1:52

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}} $.

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thanks a lot, very helpful indeed. – Carlo Beenakker Nov 6 '12 at 19:33
@Garth Payne So in the light of your answer does this function $\prod _{n=1} ^{\infty} (1+x^n)$ have any modular properties? Also can you give a more specific reference to this Euler's theorem that you refer to whose special this is you say. – Anirbit Nov 21 '12 at 3:20
@anirbit. Euler's theorem is in George Andrew's book The Theory of Partitions (p. 5). The proof I refer to above is given there in more detail. In case you don't have access to the book, someone has dupiicated the proof here:… . – Garth Payne Nov 27 '12 at 15:22

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