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In this question, there's a quite simple closed form mentioned for the series $$\sum_{n=1}^{\infty} \frac{1}{n(e^{2\pi n}-1)}$$

However I'm wondering if there exists any simple closed form known for the series of type

$$\sum_{n=1}^{\infty} \frac{1}{n(e^{\pi n}+1)}, \quad \sum_{n=1}^{\infty} \frac{1}{n(e^{2\pi n}+1)},\quad \sum_{n=1}^{\infty} \frac{1}{n(e^{(2k+1)n\pi }+1)},\quad \sum_{n=1}^{\infty} \frac{1}{n(e^{(2k)n\pi}+1)}$$

I've been thinking about thee kinds of series from quite some time and so far I have tried contour integration, cotangent partial fraction, and applying Poisson summation or converting to an integral but so far no success. Any help would be highly appreciated!

Paramanand Singh
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Permutator
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  • @pisco Of course a natural number. – Permutator Jul 09 '21 at 18:36
  • @pisco Fixed, thanks – Permutator Jul 09 '21 at 18:44
  • I think the first sum is equal to $$\frac{\pi}{8}+\ln\Gamma\left(\frac{1}{4}\right)-\frac{9}{8}\ln2-\frac{3}{4}\ln \pi$$ See https://www.wolframalpha.com/input/?i=sum%281%2F%28n%28e%5E%7BPi+n%7D%2B1%29%29%2C%7Bn%2C1%2Cinfnity%7D%29-%28Pi%2F8%2Bln%28Gamma%281%2F4%29%29-%289%2F8%29ln%282%29-3%2F4+ln%28Pi%29%29 – Omran Kouba Jul 09 '21 at 19:01
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    All your four sums can be expressed in $\pi, \Gamma(1/4)$ and some algebraic numbers (depending on $k$), it will be more complicated as $k$ increases. – pisco Jul 09 '21 at 19:01
  • And you get the second sum here https://www.wolframalpha.com/input/?i=sum%281%2F%28n%28e%5E%7B2Pi+n%7D%2B1%29%29%2C%7Bn%2C1%2Cinfnity%7D%29-%28Pi%2F4%2Bln%28Gamma%281%2F4%29%29-%287%2F4%29ln%282%29-3%2F4+ln%28Pi%29%29 – Omran Kouba Jul 09 '21 at 19:15
  • All these sums can be easily handled using the theory of theta functions and elliptic integrals. Consider $f(q) =\sum_{n\geq 1}\frac{q^{2n}}{n(1-q^{2n})}$. Then you can express all these sums in terms of $f(e^{-\pi}) $. – Paramanand Singh Jul 10 '21 at 03:19
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    For the first sum just note that it equals $f(q^{1/2})-2f(q)$ and the second sum is $f(q) - 2f(q^2)$. – Paramanand Singh Jul 10 '21 at 03:24
  • Similarly you can express the more general sums at the end of your question as $f(q^{k/2})-2f(q^k)$ where $k$ is a positive integer. – Paramanand Singh Jul 10 '21 at 03:28
  • See this answer which evaluates $f(q) $ in a closed form. – Paramanand Singh Jul 10 '21 at 03:31
  • @ParamanandSingh Can you prove that the first and the second sum equal what wolfram has provided? – Permutator Jul 10 '21 at 04:09
  • Yes, that should not be difficult given my linked answer. The sum mentioned by @OmranKouba is specifically $a(q) - 2a(q^2)$ where the values of $a(q), a(q^2)$ are given in my linked answer. Perform the calculations and you should get the desired result. – Paramanand Singh Jul 10 '21 at 06:10

1 Answers1

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With help of Mathematica I have:

$$\sum _{n=1}^{\infty } \frac{1}{n \left(e^{a n \pi }-1\right) }=\\\sum _{n=1}^{\infty } \left(\sum _{m=1}^{\infty } \frac{e^{-a n m \pi }}{n}\right)=\\\sum _{m=1}^{\infty } \left(\sum _{n=1}^{\infty } \frac{e^{-a n m \pi }}{n}\right)=\\\sum _{m=1}^{\infty } -\ln \left(1-e^{-a m \pi }\right)=\\-\frac{1}{24} (a \pi )-\ln \left(\eta \left(\frac{i a}{2}\right)\right)$$ and $$\sum _{n=1}^{\infty } \frac{1}{n (\exp (a \pi n)+1)}=\frac{a \pi }{8}-\ln \left(\eta \left(\frac{i a}{2}\right)\right)+2 \ln (\eta (i a))$$

where $a>0$ and $\eta \left(\frac{i a}{2}\right)$ is: Dedekind eta modular elliptic function.

MMA code:

Sum[1/(n*(Exp[a Pi n] - 1)), {n, 1, Infinity}] == -((a \[Pi])/24) - Log[DedekindEta[(I a)/2]]

Sum[1/(n*(Exp[a Pi n] + 1)), {n, 1, Infinity}] == (a \[Pi])/8 - Log[DedekindEta[(I a)/2]] + 2 Log[DedekindEta[I a]]

Mariusz Iwaniuk
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    Luckily the values of $\eta(ia) $ for many integer values of $a$ are already calculated on mathse:https://math.stackexchange.com/q/3101157/72031 – Paramanand Singh Jul 11 '21 at 03:31