Closed form for $\sum_{n=1}^\infty \frac{1}{n(e^{2 n \pi}-1)}$

Question

I need to calculate the sum $$\sum_{n=1}^\infty \frac{1}{n(e^{2 n \pi}-1)}$$

Write $$S=\sum_{n=1}^\infty \frac{1}{n(e^{2 n \pi}-1)}$$ $$S=\sum_{n=1}^\infty\frac{1}{n} \sum_{m=1}^\infty e^{-2 \pi n m}$$ Now changing the order of summation $$S=\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{e^{-2 \pi n m}}{n}$$$$S=-\sum_{m=1}^\infty \log(1-e^{-2\pi m})$$ So $$S=\log\left(\prod_{n=1}^\infty\frac{1}{1-e^{-2\pi n}}\right)$$ Any help would be appreciated. Thank you.

Answer 1

(More an extended comment but links to the general case.) Just in case the OP is curious about other exponents $p$, I asked a variation of it in a 2012 post which was based on Simon Plouffe's 1998 paper, "Identities Inspired from Ramanujan’s Notebooks". For the special case $p=4m+3$, Plouffe and later Joerg Arndt found,

$$\sum_{n= 1}^\infty \frac{1}{n^3(e^{2\pi n}-1)}=\frac{7}{360}\pi^3-\frac12\zeta(3)$$ $$\sum_{n= 1}^\infty \frac{1}{n^7(e^{2\pi n}-1)}=\frac{19}{113400}\pi^7-\frac12\zeta(7)$$ $$\sum_{n= 1}^\infty \frac{1}{n^{11}(e^{2\pi n}-1)}=\frac{1453}{851350500}\pi^{11}-\frac12\zeta(11)$$

and so on. The general formula for $\zeta(2n+1)$ with odd $\color{red}n$, hence $\zeta(4m+3)$, can be found in the post linked by both Singh and Manzoni.

However, the OP's exponent uses the form $p = 4m+1$ so has different results, namely,

$$\log(2) = -\frac{4}{3}\sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}-1)} -\frac{4}{3}\sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}+1)}+\frac{2}{9}\pi$$

$$\zeta(5) = -\frac{72}{35}\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}-1)} -\frac{2}{35}\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}+1)}+\frac{1}{294}\pi^5$$

$$\zeta(9) = -\frac{992}{495}\sum_{n= 1}^\infty \frac{1}{n^9(e^{2\pi n}-1)} -\frac{2}{495}\sum_{n= 1}^\infty \frac{1}{n^9(e^{2\pi n}+1)}+\frac{125}{3704778}\pi^9$$

with the last two by Plouffe and Arndt, and so on. Presumably, there should also be a concise general formula for $\zeta(4m+1)$.

---

Note 1: In the 2012 post I cited, Marko Riedel uses a general approach that handles both $p=4m\pm1$, but didn't give an explicit formula.

Note 2: Combining known results for $p=1$, we have the nice,

\begin{align} \sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}-1)} &=+\frac {3\log(\pi)}4-\log\Big(\Gamma\big(\tfrac 14\big)\Big)-\frac {\pi}{12}+\log(2)\\ \sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}+1)} &=-\frac{ 3\log(\pi)}4+\log\Big(\Gamma\big(\tfrac 14\big)\Big)\,+\,\frac {\pi}{4}-\frac{7\log(2)}4 \end{align}

Answer 2

Your last expression is in fact $\;\displaystyle S=-\log \phi\left(e^{-2\pi}\right)\,$ with $\phi$ the Euler function $\;\displaystyle \phi(q):=\prod_{n=1}^\infty 1-q^n$.

Fortunately Ramanujan found that $\;\displaystyle \phi\left(e^{-2\pi}\right)=\frac {e^{\,\pi/12}\,\Gamma\left(\frac 14\right)}{2\pi^{3/4}}\,$
(proof by Bruce Berndt in page $326$ of Ramanujan's Notebooks Part V )

allowing us some simplification of Paramanand Singh's powerful general answer :

$$\boxed{S=\log(2)+\frac 34\log(\pi)-\frac {\pi}{12}-\log\left(\Gamma\left(\frac 14\right)\right)}$$

Answer 3

Tito Piezas III added (not long ago...) : $$\zeta(5) = -\frac{72}{35}\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}-1)} -\frac{2}{35}\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}+1)}+\frac{1}{294}\pi^5$$

$$\zeta(9) = -\frac{992}{495}\sum_{n= 1}^\infty \frac{1}{n^9(e^{2\pi n}-1)} -\frac{2}{495}\sum_{n= 1}^\infty \frac{1}{n^9(e^{2\pi n}+1)}+\frac{125}{3704778}\pi^9$$

"Presumably, there should also be a concise general formula for $\zeta(4+1)$."

From this challenge (and the next solutions up to $4m+1=49$) I'll conjecture a general formula for $m>0$ :

$$\zeta(4m+1)-q_{4m+1}\,\pi^{4m+1}=\frac {-2}{2^{2m}\left(2^{2m+1}-(-1)^m\right)-1}\left(2^{2m}\left(2^{2m+1}-(-1)^m\right)\sum_{n= 1}^\infty \frac{1}{n^{4m+1}(e^{2\pi n}-1)}+\sum_{n= 1}^\infty \frac{1}{n^{4m+1}(e^{2\pi n}+1)}\right)$$ I didn't get an expression for $\,q_m\in\mathbb{Q}\,$ yet but would conjecture that it could be similar to the Bernoulli sums from this answer.

ADDITION:
Following expression for $\zeta(4m+1)$ was indeed provided by Linas Vepstas in "On Plouffe's Ramanujan Identities" (from Zander's answer) : \begin{align}&\left(1+(-4)^{m}-2^{4m+1}\right)\zeta(4m+1)\qquad\qquad\qquad\qquad\qquad\\ &=2\sum_{n=1}^{\infty}\frac{1}{n^{4m+1}\left(e^{2\pi n}+1\right)}+2\left(2^{4m+1}-(-4)^{m}\right)\sum_{n=1}^{\infty}\frac{1}{n^{4m+1}\left(e^{2\pi n}-1\right)}\\ &+(2\pi)^{4m+1}\left(\sum_{j=0}^{m}(-4)^{m+j}\frac{B_{4m-4j+2}}{(4m-4j+2)!}\frac{B_{4j}}{(4j)!}+\sum_{j=0}^{2m+1}\frac{(-4)^{j}}2\frac{B_{4m-2j+2}}{(4m-2j+2)!}\frac{B_{2j}}{(2j)!}\right) \end{align}

Mathworld contains the first odd zeta expressions with $\displaystyle S_{\pm}(n):=\sum_{k=1}^\infty \frac{1}{k^n(e^{2 \pi k}\pm 1)}$ : \begin{align} \zeta(3)&=\frac 7{180}\pi^3 - 2S_-(3)\\ \zeta(5)&=\frac 1{294}\pi^5 -\frac {72}{35} S_-(5)-\frac 2{35}S_+(5)\\ \zeta(7)&=\frac {19}{56700}\pi^7 - 2S_-(7)\\ \zeta(9)&=\frac {125}{3704778}\pi^9 -\frac {992}{495} S_-(9)-\frac 2{495}S_+(9)\\ \zeta(11)&=\frac {1453}{425675250}\pi^{11} - 2S_-(11)\\ \zeta(13)&=\frac {89}{257432175}\pi^{13} -\frac {16512}{8255} S_-(13)-\frac 2{8255}S_+(13)\\ \end{align}

To emphasize the similarities between these lines let's observe :

• the $\zeta(4m+1)$ terms could be rewritten as $\displaystyle -2S_-(4m+1)+\frac 2{1+(-4)^{m}-2^{4m+1}}\left(S_-(4m+1)+S_+(4m+1)\right)$

• the ratios $\zeta(2m+1)$ divided by $q_{2m+1}\;\pi^{2m+1}$ converge quickly to $\;\tanh(\pi)\approx 0.99627207622$