2
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Can you prove this identity?

$$\exp \left(\sum _{n=1}^{\infty} \frac{1}{n \left(1-c^n\right)}\right)=\prod _{n=1}^{\infty} \left(1-\frac{1}{c^n}\right)$$

where $|c|>2$.

I found this conjecture by entering the expression similar to the expression in the left hand side: $$\sum _{n=1}^{\infty} \frac{1}{n \left(1-2^n\right)}$$ into Excel and looking up the number in the advanced version of the inverse symbolic calculator which told me that:

$$\exp \left(\sum _{n=1}^{\infty} \frac{1}{n \left(1-2^n\right)}\right)=\prod _{n=1}^{\infty} \left(1-\frac{1}{2^n}\right)$$

After that I generalized by guessing that $2$ could be replaced by $c$.

Update 25.1.2017:

Let $$s=0 \text{ or } s=1$$ $$\sum _{n=1}^{\infty} \frac{1}{\left(1-c^n\right) n^s}=\sum _{n=1}^{\infty} \lim_{z\to s} \, (z-2) \zeta (z) \left(1-\frac{1}{\left(\frac{c^n}{c^n-1}\right)^{z-1}}\right)$$

Associated Mathematica code of these two small attempts to generalize as above:

(*Mathematica start*)
Clear[s, z, n]
s = 0;
c = N[2, 50];
Sum[1/(n^s*(1 - c^n)), {n, 1, 200}]
Sum[Limit[(z - 2)*Zeta[z]*(1 - 1/(c^n/(c^n - 1))^(z - 1)), 
  z -> s], {n, 1, 100}]

s = 1;
c = N[2, 50];
Sum[1/(n^s*(1 - c^n)), {n, 1, 200}]
Sum[Limit[(z - 2)*Zeta[z]*(1 - 1/(c^n/(c^n - 1))^(z - 1)), 
  z -> s], {n, 1, 100}]
(*end*)

-1.6066951524152917637833015231909245804805796715058

-1.606695152415291763783301523190135719575358660

-1.2420620948124149457978454818946296689734039782504

-1.242062094812414945797845481893840808068182966

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  • 1
    $\begingroup$ First replace $1/c$ with $q$ and then take logs. You need to show that $$\log\prod(1 - q^{n}) = -\sum\frac{q^{n}}{n(1 - q^{n})}$$ which is easily done by noting that $$\log\prod(1 - q^{n}) = \sum\log(1 - q^{n}) = -\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\frac{q^{mn}}{m}$$ and interchange order of summation to get the desired result. $\endgroup$ – Paramanand Singh Sep 7 '16 at 9:42
  • $\begingroup$ From my previous comment you can see that the identity holds for $|q| < 1$ i.e. $|c| > 1$. $\endgroup$ – Paramanand Singh Sep 7 '16 at 11:11
  • 1
    $\begingroup$ The expression $$a(q) = \sum\frac{q^{n}}{n(1 - q^{n})}$$ has a closed form involving elliptic integrals and there is a nice formula for $\pi$ in terms of $a(q)$. See math.stackexchange.com/q/938123/72031 for more details. $\endgroup$ – Paramanand Singh Sep 7 '16 at 11:17
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    $\begingroup$ @ParamanandSingh it is amazing seeing MatsGranvik (almost) discovering his own modular form on Excel, when Poincarre took several years to do the same :D $\endgroup$ – reuns Sep 7 '16 at 11:23
  • $\begingroup$ @user1952009: modern times have lead to modern tools like Excel! Hopefully these newer tools will help to discover many new thing more easily. $\endgroup$ – Paramanand Singh Sep 7 '16 at 16:03
-1
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Possible hint.

$$e^{\sum_{n=1}^{\infty}\frac{1}{n \left(1-c^n\right)}} = e^{\frac{1}{1\cdot \left(1-c^1\right)} + \frac{1}{2\cdot \left(1-c^2\right)} + \frac{1}{3\cdot \left(1-c^3\right)} + \ldots} = e^{\frac{1}{1\cdot \left(1-c^1\right)}} e^{\frac{1}{2\cdot \left(1-c^2\right)}} e^{\frac{1}{3\cdot \left(1-c^3\right)}} e^{\ldots} = \prod_n e^{\frac{1}{n\cdot \left(1-c^n\right)}}$$

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  • $\begingroup$ How does this help? You have only used the property of exponential function $\exp(x + y) = \exp(x)\exp(y)$. The identity in question requires manipulation of double series. Moreover note that your last equality is wrong. $\endgroup$ – Paramanand Singh Sep 7 '16 at 9:38
  • $\begingroup$ @ParamanandSingh it's not wrong. And that was a hint, not the whole answer. $\endgroup$ – Von Neumann Sep 7 '16 at 9:48
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    $\begingroup$ I fail to see how this is helpful as a hint. It has nothing to do with the problem asked. $\endgroup$ – Paramanand Singh Sep 7 '16 at 9:54
  • $\begingroup$ You should probably see my comment to question when I mention the crux of the problem. $\endgroup$ – Paramanand Singh Sep 7 '16 at 10:01

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