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Ramanujan found the following formula:

$$\large \sum_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$$

I let $e^{2\pi n}-1=\left(e^{\pi n}+1\right)\left(e^{\pi n}-1\right)$ to try partial fraction decomposition and turn the sum into telescoping, but methinks it doesn't lead anywhere and only makes things hairy.

How does one go about proving this? Thanks.

Mr Pie
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    I'm pretty sure complex analysis (residues) is the only way here – Yuriy S Feb 23 '20 at 14:08
  • @YuriyS it appears that my curiosity has taken me to the deep end in maths once again - and I don't swim at the deep end :/ – Mr Pie Feb 23 '20 at 14:09
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    @YuriyS: one should not underestimate the power of basic mathematics. Ramanujan proved this identity and a lot more using basic algebraic manipulation and calculus. – Paramanand Singh Feb 24 '20 at 11:14
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    Also contrary to popular belief, Ramanujan not only found formulas he also proved them. For lack of paper he omitted many of such marvelous proofs. – Paramanand Singh Feb 24 '20 at 11:47
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    @MarkusScheuer oh so kind :) – Mr Pie Feb 29 '20 at 01:17
  • @ParamanandSingh I thought he wrote most of his proofs on the sand in a temple near where he lived, and then only wrote the formulae in the notebooks or something like that. Only one who is willing to go to such lengths to express their heart and mind like that will achieve the greatest of things. – Mr Pie Mar 01 '20 at 06:57
  • Well, Ramanujan used to write on whatever surface was available to him. This included floor of his house, slate and paper. Out of all this paper was too costly for him and he used it to record his precious formulae. There was simply not enough paper to record his proofs. – Paramanand Singh Mar 01 '20 at 07:02
  • Curious: $\sum_{n=1}^{\infty} \frac{n^{13}}{e^{2 \pi n}-1}=\int_0^{\infty } \frac{n^{13}}{\exp (2 \pi n)-1} , dn$. This holds for all powers $n^{4k+1}, k=1,2,...$ . – Dr. Wolfgang Hintze Mar 19 '21 at 18:25

6 Answers6

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Suppose we seek to evaluate

$$S = \sum_{n\ge 1} \frac{n^{13}}{e^{2\pi n}-1}.$$

This sum may be evaluated using harmonic summation techniques.

Introduce the sum $$S(x; p) = \sum_{n\ge 1} \frac{n^{4p+1}}{e^{nx}-1}$$ with $p$ a positive integer and $x\gt 0.$

The sum term is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = k^{4p+1}, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{1}{e^x-1}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{1}{e^{x}-1} x^{s-1} dx = \int_0^\infty \frac{e^{-x}}{1-e^{-x}} x^{s-1} dx \\ = \int_0^\infty \sum_{q\ge 1} e^{-q x} x^{s-1} dx = \sum_{q\ge 1} \int_0^\infty e^{-q x} x^{s-1} dx \\= \Gamma(s) \sum_{q\ge 1} \frac{1}{q^s} = \Gamma(s) \zeta(s).$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x,p)$ is given by

$$Q(s) = \Gamma(s) \zeta(s) \zeta(s-(4p+1)) \\ \text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} k^{4p+1} \frac{1}{k^s} = \zeta(s-(4p+1))$$ for $\Re(s) > 4p+2.$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{4p+5/2-i\infty}^{4p+5/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.

The two zeta function terms cancel the poles of the gamma function term and we are left with just

$$\begin{align} \mathrm{Res}(Q(s)/x^s; s=4p+2) & = \Gamma(4p+2) \zeta(4p+2) / x^{4p+2} \quad\text{and}\\ \mathrm{Res}(Q(s)/x^s; s=0) & = \zeta(0) \zeta(-(4p+1)). \end{align}$$

Computing these residues we get

$$(4p+1)! \frac{B_{4p+2} (2\pi)^{4p+2}}{2(4p+2)! \times x^{4p+2}} = \frac{B_{4p+2} (2\pi)^{4p+2}}{2\times (4p+2) \times x^{4p+2}}$$ and $$- \frac{1}{2} \times -\frac{B_{4p+2}}{4p+2}.$$

This shows that $$S(x;p) = \frac{B_{4p+2} (2\pi)^{4p+2}}{(8p+4)\times x^{4p+2}} + \frac{B_{4p+2}}{8p+4} + \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} Q(s)/x^s ds.$$

To treat the integral recall the duplication formula of the gamma function: $$\Gamma(s) = \frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right).$$

which yields for $Q(s)$

$$\frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right) \zeta(s) \zeta(s-(4p+1))$$

Furthermore observe the following variant of the functional equation of the Riemann zeta function: $$\Gamma\left(\frac{s}{2}\right)\zeta(s) = \pi^{s-1/2} \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s)$$

which gives for $Q(s)$ $$\frac{1}{\sqrt\pi} 2^{s-1} \pi^{s-1/2} \Gamma\left(\frac{s+1}{2}\right) \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s)\zeta(s-(4p+1)) \\ = \frac{1}{\sqrt\pi} 2^{s-1} \pi^{s-1/2} \frac{\pi}{\sin(\pi(s+1)/2)} \zeta(1-s)\zeta(s-(4p+1)) \\ = 2^{s-1} \frac{\pi^s}{\sin(\pi(s+1)/2)} \zeta(1-s)\zeta(s-(4p+1)).$$

Now put $s=4p+2-u$ in the remainder integral to get

$$- \frac{1}{x^{4p+2}} \frac{1}{2\pi i} \int_{4p+5/2+i\infty}^{4p+5/2-i\infty} 2^{4p+1-u} \\ \times \frac{\pi^{4p+2-u}}{\sin(\pi(4p+3-u)/2)} \zeta(u-(4p+1))\zeta(1-u) x^u du \\ = \frac{2^{4p+2} \pi^{4p+2}}{x^{4p+2}} \frac{1}{2\pi i} \int_{4p+5/2-i\infty}^{4p+5/2+i\infty} 2^{u-1} \\ \times \frac{\pi^{u}}{\sin(\pi(4p+3-u)/2)} \zeta(u-(4p+1))\zeta(1-u) (x/\pi^2/2^2)^u du.$$

Now $$\sin(\pi(4p+3-u)/2) = \sin(\pi(1-u)/2+\pi (2p+1)) \\ = - \sin(\pi(1-u)/2) = \sin(\pi(-1-u)/2) = - \sin(\pi(u+1)/2).$$

We have shown that $$\bbox[5px,border:2px solid #00A000] {S(x;p) = \frac{B_{4p+2} (2\pi)^{4p+2}}{(8p+4)\times x^{4p+2}} + \frac{B_{4p+2}}{8p+4} - \frac{(2\pi)^{4p+2}}{x^{4p+2}} S(4\pi^2/x;p)}.$$

In particular we get

$$S(2\pi; p) = \frac{B_{4p+2}}{8p+4}.$$

The sequence in $p$ starting from $p=1$ is

$${\frac{1}{504}},{\frac{1}{264}},1/24, {\frac{43867}{28728}},{\frac{77683}{552}}, {\frac{657931}{24}},{\frac{1723168255201}{171864}}, \ldots$$

We thus have for $p=3$ as per request from OP

$$\bbox[5px,border:2px solid #00A000]{ \sum_{n\ge 1} \frac{n^{13}}{e^{2\pi n}-1} = \frac{1}{24}.}$$

References, as per request, are: Flajolet and Sedgewick, Mellin transform asymptotics, INRIA RR 2956 and Szpankowski, Mellin Transform and its applications, from Average Case Analysis of Algorithms on Sequences.

Marko Riedel
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It is the weight $14$ Eisenstein series $$G_{14}(z)=\sum_{(n,m)\ne (0,0)} \frac1{(zn+m)^{14}}= 2\zeta(14)+\sum_{n\ne 0} \frac{1}{13!} \frac{d^{13}}{dz^{13}}\frac{2i\pi}{e^{2i\pi n z}-1}$$ $$=2\zeta(14)+\sum_{n\ge 1} \frac{4i\pi}{13!} \sum_{m\ge 1} (2i\pi m)^{13}e^{2i\pi mz}=2\zeta(14)+(2i\pi)^{14}\frac{2}{13!}\sum_{k\ne 1}\frac{k^{13}}{e^{-2i\pi kz}-1} $$

$$G_{14}(z)= z^{-14}G_{14}(-1/z)\implies \qquad G_{14}(i)=0$$

$$\boxed{(2i\pi)^{14}\frac{2}{13!}\sum_{k\ne 1}\frac{k^{13}}{e^{2\pi kz}-1}+2\zeta(14)=0 }$$ $2\zeta(14)=-\frac{B_{14}(2\pi)^{14}}{(14)!} $

reuns
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For your curiosity !

I do not know if these results are known but, beside this one, $$ \sum_{n=1}^\infty \frac{n^{5}}{e^{2\pi n}-1}=\frac 1{504}=\frac 1{21 \times 24}\qquad\text{and} \qquad \sum_{n=1}^\infty \frac{n^{9}}{e^{2\pi n}-1}=\frac 1{264}=\frac 1{11 \times 24}$$

If they are known, please tell me where I could find them.

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Theorem 1. (see [1] pg.275-276) Let $a,b>0$ with $ab=\pi^2$, and let $\nu$ be any non zero integer. Then
$$ a^{-\nu}\left\{\frac{1}{2}\zeta(2\nu+1)+\sum^{\infty}_{n=1}\frac{n^{-2\nu-1}}{e^{2an}-1}\right\}- (-b)^{-\nu}\left\{\frac{1}{2}\zeta(2\nu+1)+\sum^{\infty}_{n=1}\frac{n^{-2\nu-1}}{e^{2bn}-1}\right\}= $$ \begin{equation} =-2^{2\nu}\sum^{\nu+1}_{n=0}(-1)^n\frac{B_{2n}}{(2n)!}\frac{B_{2\nu+2-2n}}{(2\nu+2-2n)!}a^{\nu+1-n}b^n,\tag 1 \end{equation} where $\zeta(s)$ is the Riemann zeta function and $B_n$ is the $n-$th Bernoulli number.

Notes

For integer $\nu<-1$ formula (1) evaluated from:

Theorem 2. (see [1] pg.261) If $\nu$ is integer greater than 1, then ($ab=\pi^2$, $a,b>0$) $$ a^{\nu}\sum^{\infty}_{n=1}\frac{n^{2\nu-1}}{e^{2an}-1}-(-b)^{\nu}\sum^{\infty}_{n=1}\frac{n^{2\nu-1}}{e^{2bn}-1}=(a^{\nu}-(-b)^{\nu})\frac{B_{2\nu}}{4\nu}\tag 2 $$

[1]: B.C. Berndt, 'Ramanujan`s Notebooks Part II'. Springer Verlang, New York., (1989).

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    This formula of Ramanujan is based on another related formula which is being discussed in this thread. – Paramanand Singh Mar 01 '20 at 07:10
  • Also while Berndt has done an excellent job proving Ramanujan's results, his work is based on a lot of references. Because of this I prefer online content where references are available with a click. – Paramanand Singh Mar 01 '20 at 07:13
  • BTW how do you interpret this formula when $\nu$ is negative? Does the sum containing Bernoulli stuff is supposed to be equal to $0$ when $\nu$ is negative? – Paramanand Singh Mar 01 '20 at 07:31
  • I think both your results can be provided by theorem 1 if we adopt the convention that for $\nu\leq - 1$ the sum on right is $0$ and $\zeta(-m) =-\dfrac{B_{m+1}}{m+1}$ where $m$ is odd positive integer. – Paramanand Singh Mar 02 '20 at 04:08
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Essentially the same as @reuns answer but with more detail.

Let $k\ge2$ be an integer and define the Eisenstein series $$G_{2k}(z)=\sum_{(n,m)\in A}\frac{1}{(n+mz)^{2k}},\tag 1$$ where $A=\Bbb Z^2\setminus\{(0,0)\}$, and $z\in\Bbb C$ with $\Im(z)>0$. It is simple to show that $G_{2k}(z+1)=G_{2k}(z)$ for all $z$, so it follows that we may write $$G_{2k}(z)=\sum_{n\ge0}g_nq^n,$$ where $q=e^{2i\pi z}$. It can be shown (see here) that there is a closed form for $g_n$. Namely, we can write $$\begin{align} G_{2k}(z)&=2\zeta(2k)\left(1+c_{2k}\sum_{n\ge1}\sigma_{2k-1}(n)q^n\right)\\ &=2\zeta(2k)\left(1+c_{2k}\sum_{n\ge1}\frac{n^{2k-1}q^n}{1-q^n}\right), \end{align}$$ where $c_{2k}=\frac{(2\pi i)^{2k}}{(2k-1)!\zeta(2k)}=\frac{-4k}{B_{2k}}=\frac{2}{\zeta(1-2k)}$.

On the other hand, it is simple to show from $(1)$ that $$G_{2k}(-1/z)=z^{2k}G_{2k}(z).$$ Letting $E_{2k}(z)=\frac{1}{2\zeta(2k)}G_{2k}(z)$ for convenience, we have $$E_{2k}(-1/z)=z^{2k}E_{2k}(z).\tag2$$ Then defining $$S_k(q)=\sum_{n\ge1}\frac{n^{2k-1}}{q^n-1},$$ we have $$E_{2k}(z)=1+c_{2k}S_k(e^{-2i\pi z}).$$ Then from $(2)$, we have $$1+c_{2k}S_k(e^{2i\pi/z})=z^{2k}\left(1+c_{2k}S_k(e^{-2i\pi z})\right).\tag3$$ Since your sum is given by $S_7(e^{2\pi})$, we set $k=7$ and $z=i$ in $(3)$, and get $$\begin{align} 1+c_{14}S_7(e^{2\pi})&=-\left(1+c_{14}S_7(e^{2\pi})\right)\\ \Rightarrow S_7(e^{2\pi})&=-\frac{1}{c_{14}}. \end{align}$$ Since $c_{2k}=-4k/B_{2k}$, we have $-1/c_{14}=B_{14}/28=1/24$, and thus $$S_7(e^{2\pi})=\sum_{n\ge1}\frac{n^{13}}{e^{2\pi n}-1}=\frac{1}{24}.$$ In general, the same method allows us to compute $$S_{2k+1}(e^{2\pi})=\sum_{n\ge1}\frac{n^{4k+1}}{e^{2\pi n}-1}=\frac{B_{4k+2}}{8k+4},$$ as was shown by @MarcoRiedel.

clathratus
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This is not a strict solution but a heuristic proof using CAS. It also shows that there are other "magic" numbers instead of 13 which result in similar simple fractions. Playing around I also found the interesting conincidence that in some cases the sum and the related integral give the same results.

Defining

$$S(m) = \sum_{n=1}^{\infty} \frac{n^m}{e^{2 \pi n}-1}\tag{1}$$

we have to show that $S(13) = \frac{1}{24}$.

My first idea was to identify the denominator. I knew this expression from the Planck radioation formula and from the generating function of the Benoulli numbers.

But why so complicated? It is just the sum of a power series. Indeed we can write

$$\frac{1}{e^{2 \pi n}-1} = \sum _{j=1}^{\infty } \exp (- 2 \pi j n)\tag{2}$$

Next, for the numerator we replace the power of $n$ by $z^n$ and consider the intermediate generating function

$$g_{0}(z,j) = \sum _{n=1}^{\infty } z^n \exp (- 2 \pi j n)=\frac{z}{e^{2 \pi j}-z}\tag{3}$$

Summing over $j$ gives the generating function

$$\begin{align}g(z) = & \sum _{j=1}^{\infty } \frac{z}{e^{2 \pi j}-z}\\=& -\frac{1}{2 \pi}\psi _{e^{-2 \pi }}^{(0)}\left(1-\frac{\log (z)}{2 \pi }\right)+\frac{1}{2 \pi} \log \left(1-e^{-2 \pi }\right) \end{align}\tag{4}$$

Here $\psi _{q }^{(0)}(x)$ is the q-digamma function.

The the sums in question here can be found as derivatives of $g(z)$

$$S(m,z) = (z\frac{\partial}{\partial z})^m g(z) |_{z \to 1}\tag{5}$$

Using Mathematica for some values of $m$ I saw the pattern and found the general formula

$$S(m,z)=\frac{(-1)^{m+1} \psi _{e^{-2 \pi }}^{(m)}\left(1-\frac{\log (z)}{2 \pi }\right)}{2^{m+1} \pi ^{m+1}}\tag{6}$$

which gives at $z=1$ where the $\log$ vanishes the result

$$S(m)=S(m,z\to 1) = \frac{1}{(2 \pi )^{n+1}} \psi _{e^{-2 \pi }}^{(n)}(1)\tag{7}$$

Finally, the numerical results in Mathematica for $\frac{1}{S(m)}$ were particularly simple for 3 values of $m$. In the format {m,1/S(m)} we have

$$\{\{5, 504\}, \{9, 264\}, \{13,24\}\}\tag{8}$$

The case $m=13$ leads to the desired result. The "magic" values of $m$ have the form $1+4k$.

Discussion

It is always tempting with sums to look at the corresponding integral.

In our case we consider

$$i(m) = \int_0^{\infty } \frac{n^m}{\exp (2 \pi n)-1} \, dn\tag{9}$$

where in contrast to the sum the integration starts at $n=0$.

The surprising observation is that we have (checked numerically)

$$i(m) = S(m),m=1+4k, k=1,2,...\tag{10}$$

which includes the "magic" values.

It would be nice to find a proof of this observation

  • A bit too late ! This is very interesting : we have $$i(m)=(2 \pi )^{-(m+1)} \text{Li}_{m+1}(1) \Gamma (m+1)$$ What a coincidence ! As you wrote, "It would be nice to find a proof of this observation". Cheers :-) – Claude Leibovici Apr 08 '21 at 10:07
  • Claude, nice things are never "too late" ;-). Have you seen this https://math.stackexchange.com/q/4070922/198592? – Dr. Wolfgang Hintze Apr 08 '21 at 13:15