22
$\begingroup$

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.

$\endgroup$
9
  • 4
    $\begingroup$ I'm pretty sure complex analysis (residues) is the only way here $\endgroup$
    – Yuriy S
    Feb 23 '20 at 14:08
  • 4
    $\begingroup$ @YuriyS: one should not underestimate the power of basic mathematics. Ramanujan proved this identity and a lot more using basic algebraic manipulation and calculus. $\endgroup$ Feb 24 '20 at 11:14
  • 2
    $\begingroup$ See this answer. $\endgroup$ Feb 24 '20 at 11:43
  • 3
    $\begingroup$ 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. $\endgroup$ Feb 24 '20 at 11:47
  • 1
    $\begingroup$ @MarkusScheuer oh so kind :) $\endgroup$
    – Mr Pie
    Feb 29 '20 at 1:17
25
+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.

$\endgroup$
7
  • $\begingroup$ Great answer (+1) Would you mind to recommend a reference to study this and related problems? $\endgroup$
    – epi163sqrt
    Feb 23 '20 at 19:52
  • $\begingroup$ Thank you and references added. $\endgroup$ Feb 23 '20 at 21:41
  • 1
    $\begingroup$ Many thanks for this interesting information. $\endgroup$
    – epi163sqrt
    Feb 23 '20 at 21:42
  • 2
    $\begingroup$ This is fantastic answer ! Thanks $\endgroup$ Feb 24 '20 at 3:40
  • $\begingroup$ Thanks muchly, and congratulations! $(+1)$ $\color{green}{\checkmark}$ :) $\endgroup$
    – Mr Pie
    Feb 24 '20 at 13:43
17
$\begingroup$

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)!} $

$\endgroup$
2
  • 5
    $\begingroup$ 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 '20 at 1:48
  • $\begingroup$ Didn't you mean $k\ne 0$ im the last two sums? $\endgroup$ Mar 24 at 18:45
12
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ This is amazing! It appears that there is a closed form for every $n^{4k+1}$. $\endgroup$
    – Mr Pie
    Feb 23 '20 at 22:46
6
$\begingroup$

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).

$\endgroup$
4
  • 1
    $\begingroup$ This formula of Ramanujan is based on another related formula which is being discussed in this thread. $\endgroup$ Mar 1 '20 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. $\endgroup$ Mar 1 '20 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? $\endgroup$ Mar 1 '20 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. $\endgroup$ Mar 2 '20 at 4:08
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 :-) $\endgroup$ Apr 8 at 10:07
  • $\begingroup$ Claude, nice things are never "too late" ;-). Have you seen this math.stackexchange.com/q/4070922/198592? $\endgroup$ Apr 8 at 13:15

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.