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I recently found the following result on Twitter.

$$\sum_{n=1}^\infty \frac{n^5}{e^{2\pi n}-1}=\frac{1}{504}$$

I know that $\int_0^\infty \frac{x^5}{e^{2 \pi x}-1} dx = \frac{5!}{(2\pi)^6}\zeta(6)=\frac{1}{504}$

How to show that the sum is also equal to the same number?

Can this be generalized to a class of functions where the sum (over positive integers) and the definite integral (from 0 to $\infty$) are the same?

Archisman Panigrahi
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    I suspect it is one of those very intimidating modular form identities – FShrike Oct 23 '22 at 20:19
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    Or the following questions (1), (2), and (3) to name a few. As @FShrike mentioned, sums of this form have been asked about on MSE several times in the past and are all related to modular form theory and Eisenstein series. – KStarGamer Oct 23 '22 at 20:23
  • It seems that the general case (with $\frac{n^k}{e^{2 \pi n}-1}$) is quite involved (I am not familiar with modular forms and Mellin transforms). I will wait for some time to see if I get an intuitive answer of how to evaluate this particular sum. – Archisman Panigrahi Oct 23 '22 at 20:28
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    There is no simpler way than relating it to $\sum_{(a,b)\in \Bbb{Z}^2- (0,0)} \frac1{(a+ib)^6} = 0$ using the partial fraction decomposition of $\sum_{n\ge 1} n^5 e^{2i\pi nz}$. The same identity holds for any integer $\equiv 1\bmod 4$. – reuns Oct 23 '22 at 21:58
  • With Mathematica: $$\sum {n=1}^{\infty } \frac{n^k}{\exp (2 \pi n)-1}=\sum _{m=0}^{\infty } (-1)^{-m} \cos (m \pi ) \text{Li}{-k}\left(e^{-2 (1+m) \pi }\right)=\left(-\frac{1}{2 \pi }\right)^{1+k} \psi _{e^{2 \pi }}^{(k)}(1)$$ where: $\psi _{e^{2 \pi }}^{(k)}(1)$ is the n^th derivative of the q-digamma function. – Mariusz Iwaniuk Oct 26 '22 at 09:12

2 Answers2

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I'm not sure if it's more appropriate to post this here or elsewhere, but we can integrate the function $$f(z) = \frac{z^{5}e^{iz}}{\cosh(z) - \cos(z)} $$ around the contour $$\left[-\frac{(2N+1) \pi \sqrt{2}}{2}, \frac{(2N+1) \pi \sqrt{2}}{2}\right] \cup \frac{(2N+1) \pi \sqrt{2}}{2} e^{i[0, \pi]}. $$

(For any positive integer $N$, the semicircular part of the above contour passes halfway between two adjacent poles of $f(z)$.)

As $N \to \infty$ through the positive integers, the magnitude of $f(z)$ decays exponentially to zero on the semicircle. This is due to the fact that the magnitude of $e^{iz}$ decays exponentially to zero as $\Im(z) \to + \infty$, while the magnitude of $\cosh(z)$ grows exponentially as $\Re(z) \to \pm \infty$.

We therefore have $$\begin{align} \int_{-\infty}^{\infty} f(x) \, \mathrm dx &= 2 \pi i \left(\sum_{n=1}^{\infty}\operatorname{Res}\left[f(z), n \pi(1+i) \right]+ \sum_{n=1}^{\infty}\operatorname{Res}\left[f(z), n \pi(-1+i) \right] \right) \\ &= 2 \pi i \left(\sum_{n=1}^{\infty}\lim_{z \to n\pi(1+i)} \frac{z^{5}e^{iz}}{\sinh(z) + \sin (z)} + \sum_{n=1}^{\infty} \lim_{z \to n \pi(-1+i)} \frac{z^{5}e^{iz}}{\sinh(z) + \sin (z)} \right) \\&= 2 \pi i \left(\sum_{n=1}^{\infty} \frac{n^{5} \pi^{5}(1+i)^{5}(-1)^{n} e^{- n \pi}}{(-1)^{n}(1+i)\sinh(n \pi)} + \sum_{n=1}^{\infty} \frac{n^{5} \pi^{5}(-1+i)^{5}(-1)^{n}e^{-n \pi}}{(-1)^{n}(-1+i) \sinh(n \pi)} \right) \\ &= -16 \pi^{6} i \sum_{n=1}^{\infty} \frac{n^{5}e^{-n \pi}}{\sinh(n \pi)} \\ &=-32 \pi^{6} i \sum_{n=1}^{\infty} \frac{n^{5}}{e^{2 \pi n}-1}. \end{align} $$

Equating the imaginary parts on both sides of the equation, we get $$ \begin{align} \sum_{n=1}^{\infty} \frac{n^{5}}{e^{2 \pi n}-1} &= -\frac{1}{32 \pi^{6}} \int_{-\infty}^{\infty} \frac{x^{5} \sin(x)}{\cosh(x) - \cos(x)} \, \mathrm dx \\ &= -\frac{1}{16 \pi^{6}} \int_{0}^{\infty}\frac{x^{5} \sin(x)}{\cosh(x) - \cos(x)} \, \mathrm dx \\ &= -\frac{1}{8 \pi^{6}} \, \Im \int_{0}^{\infty}x^{5} \sum_{n=1}^{\infty} e^{-(1-i)nx} \, \mathrm dx \\ &= -\frac{1}{8 \pi^{6}} \, \Im \sum_{n=1}^{\infty} \int_{0}^{\infty} x^{5} e^{-(1-i)nx} \, \mathrm dx \\ &= -\frac{1}{8 \pi^{6}} \, \Im \sum_{n=1}^{\infty} \frac{\Gamma(6)}{(1-i)^6n^{6}} \\ &= \frac{15}{8 \pi^{6}} \sum_{n=1}^{\infty} \frac{1}{n^{6}} \\ &= \frac{1}{504}. \end{align}$$

Random Variable
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\sum_{n = 1}^{\infty}{n^{5} \over \expo{2\pi n} - 1} = {\LARGE ?}}$

The ${\ds{t^{5}} \over \ds{\expo{2\pi t} - 1}}$- $\ds{Mellin\ Transform}$ is given by: \begin{align} & \left.\int_{0}^{\infty}t^{z - 1} \,{t^{5} \over \expo{2\pi t} - 1}\dd t \right\vert_{\Re\pars{z}\ >\ -4} = \pars{2\pi}^{-z - 5}\,\,\, \Gamma\pars{z + 5}\zeta\pars{z + 5} \\[5mm] & \mbox{such that} \\ & {t^{5} \over \expo{2\pi t} - 1} = \left.\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic} \pars{2\pi}^{-z - 5}\,\,\,\Gamma\pars{z + 5}\zeta\pars{z + 5}t^{-z}\,\,{\dd z \over 2\pi\ic} \right\vert_{\,\,c\ >\ -4} \end{align} Hereafter, I'll choose $\ds{c = 1^{+}}$ in order to ensure the convergence of $\ds{\sum_{n = 1}^{\infty}n^{-z}}$. Then, \begin{align} & \sum_{n = 1}^{\infty}{n^{5} \over \expo{2\pi n} - 1} \\[5mm] = & \int_{1^{+}\ -\ \infty\ic}^{1^{+}\ +\ \infty\ic}\ \overbrace{\pars{2\pi}^{-z - 5}\,\,\,\Gamma\pars{z + 5}\zeta\pars{z + 5}\zeta\pars{z}} ^{\ds{\on{f}\pars{z}}}\,\,{\dd z \over 2\pi\ic} \end{align} I'll switch the integration to the path $\ds{\braces{z \mid z = -2 + t\ic,\ t \in \mathbb{R}}}$: \begin{align} & \sum_{n = 1}^{\infty}{n^{5} \over \expo{2\pi n} - 1} = \int_{-\infty}^{\infty}\on{f}\pars{-2 + \ic t} {\dd t \over 2\pi} + \on{Res}\bracks{\on{f}\pars{z},z = 1} \end{align} The integral vanishes out: With $\ds{Riemann\ Functional\ Equation}$ it's shown that the integrand is an $\ds{odd}$ function of $\ds{t}$. Integration along $\ds{\pars{-2 \pm \ic R,1^{+} \pm \ic R}}$ vanishes out too as $\ds{R \to \infty}$.

Therefore, \begin{align} & \sum_{n = 1}^{\infty}{n^{5} \over \expo{2\pi n} - 1} = \on{Res}\bracks{\on{f}\pars{z},z = 1} = \pars{2\pi}^{-6}\,\,\Gamma\pars{6}\zeta\pars{6} \\[5mm] = & {1 \over 64\pi^{6}}\times 120 \times {\pi^{6} \over 945} = \bbx{\color{#44f}{1 \over 504}} \approx 0.001984 \end{align}

Felix Marin
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  • How do you calculate the residue of $\zeta(z)$ at $z=1$? From your answer, it seems that it is 1, but I don't know how to show it. I can follow the rest of the answer. – Archisman Panigrahi Oct 27 '22 at 16:46
  • @ArchismanPanigrahi $\zeta\left(z\right)$ has a single pole at $z = 1$ with residue igual to $1$. The multiplicative factors - i.e. $\left(2\pi\right)^{-z - 5},,,,\Gamma\left(z + 5\right)\zeta\left(z + 5\right)$- are finite at $z = 1$. – Felix Marin Nov 02 '22 at 02:36
  • I understood that. What I wanted to ask is, why is the residue of $\zeta(z)$ equal to 1 at $z=1$? Is there a way to show that it is a simple pole? – Archisman Panigrahi Nov 02 '22 at 03:59
  • @ArchismanPanigrahi It appears in every book about the $\zeta$ function. This is a limited space to discuss it. – Felix Marin Nov 03 '22 at 00:09