19
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ I'm pretty sure complex analysis (residues) is the only way here $\endgroup$ – Yuriy S Feb 23 at 14:08
  • 2
    $\begingroup$ @YuriyS: one should not underestimate the power of basic mathematics. Ramanujan proved this identity and a lot more using basic algebraic manipulation and calculus. $\endgroup$ – Paramanand Singh Feb 24 at 11:14
  • 2
    $\begingroup$ See this answer. $\endgroup$ – Paramanand Singh Feb 24 at 11:43
  • 2
    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Feb 24 at 11:47
  • 1
    $\begingroup$ @MarkusScheuer oh so kind :) $\endgroup$ – Mr Pie Feb 29 at 1:17
22
+100
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Great answer (+1) Would you mind to recommend a reference to study this and related problems? $\endgroup$ – Markus Scheuer Feb 23 at 19:52
  • $\begingroup$ Thank you and references added. $\endgroup$ – Marko Riedel Feb 23 at 21:41
  • 1
    $\begingroup$ Many thanks for this interesting information. $\endgroup$ – Markus Scheuer Feb 23 at 21:42
  • 2
    $\begingroup$ This is fantastic answer ! Thanks $\endgroup$ – Claude Leibovici Feb 24 at 3:40
  • $\begingroup$ Thanks muchly, and congratulations! $(+1)$ $\color{green}{\checkmark}$ :) $\endgroup$ – Mr Pie Feb 24 at 13:43
12
$\begingroup$

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)!} $

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ Note that evaluating $G_{4k}(i)$ is more complicated, we need to relate $G_4(i)$ to some elliptic integral then to the $\beta$ function thus to the $\Gamma$ function $\endgroup$ – reuns Feb 26 at 1:48
10
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This is amazing! It appears that there is a closed form for every $n^{4k+1}$. $\endgroup$ – Mr Pie Feb 23 at 22:46
5
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This formula of Ramanujan is based on another related formula which is being discussed in this thread. $\endgroup$ – Paramanand Singh Mar 1 at 7:10
  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh Mar 1 at 7:13
  • $\begingroup$ 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? $\endgroup$ – Paramanand Singh Mar 1 at 7:31
  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh Mar 2 at 4:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.