26
$\begingroup$

Show that

$$\sum_{n=1}^{\infty}\frac{\sinh\big(\pi n\sqrt2\big)-\sin\big(\pi n\sqrt2\big)}{n^3\Big({\cosh\big(\pi n\sqrt2}\big)-\cos\big(\pi n\sqrt2\big)\Big)}=\frac{\pi^3}{18\sqrt2}$$

I have no hint as to how to even start.

$\endgroup$
  • $\begingroup$ What is the argument of $\cosh$ in the denominator? $\endgroup$ – user64494 Oct 12 '13 at 15:40
  • $\begingroup$ @user64494 sorry I have to change $\endgroup$ – Shivam Patel Oct 12 '13 at 15:42
  • 3
    $\begingroup$ @user64494 I to admire the beauty of it..... $\endgroup$ – Shivam Patel Oct 13 '13 at 2:55
  • 4
    $\begingroup$ I think it would be helpful to provide a source for this problem. $\endgroup$ – Slade Sep 30 '14 at 5:35
  • 2
    $\begingroup$ @Slade I got it playing around with mathematica and the Inverse symbolic calculator. $\endgroup$ – Shivam Patel Oct 1 '14 at 6:45
37
+50
$\begingroup$
  1. It is a simple trigonometric exercise to show that $$\frac{\sinh\pi n\sqrt2 -\sin\pi n\sqrt2}{\cosh\pi n\sqrt2 -\cos\pi n\sqrt2}= \Re \left(\coth \pi n z_0+z_0^2\coth\frac{\pi n}{z_0}\right),\tag{$\spadesuit$}$$ with $z_0=e^{i\pi/4}$.

  2. Recall the Ramanujan identity (several nice proofs of which may be found here): $$\sum_{n=1}^{\infty}\frac{1}{n^3}\left(\coth \pi n x+x^2\coth\frac{\pi n}{x}\right)= \frac{\pi^3}{90x}\left(x^4+5x^2+1\right).\tag{$\heartsuit$}$$

  3. Combining ($\spadesuit$) and ($\heartsuit$), we immediately get $$\sum_{n=1}^{\infty}\frac1{n^3}\left(\frac{\sinh\pi n\sqrt2 -\sin\pi n\sqrt2}{\cosh\pi n\sqrt2 -\cos\pi n\sqrt2}\right)=\frac{\pi^3}{90}\,\Re\frac{z_0^4+5z_0^2+1}{z_0}=\frac{\pi^3}{18\sqrt2}.$$

$\endgroup$
  • $\begingroup$ Did not see the connection to the Ramanujan problem coming! Very nice. $\endgroup$ – Semiclassical Oct 1 '14 at 12:12
  • 6
    $\begingroup$ @Semiclassical It was obvious - the OP is a huge fan of Ramanujan ;) $\endgroup$ – Start wearing purple Oct 1 '14 at 12:14
  • $\begingroup$ True. I had been looking for a connection between this identity and solutions of some BVP. (For instance, the summands are linked with the dfn of bipolar coordinates.) $\endgroup$ – Semiclassical Oct 1 '14 at 13:52
  • 2
    $\begingroup$ You elucidated the mystery by pointing towards an enigma and showing that the two are the same. $\endgroup$ – Lucian Oct 1 '14 at 14:36
  • $\begingroup$ +1 for using my passions as a hint for such a beautiful proof... True I am a great fan of ramanujan :) $\endgroup$ – Shivam Patel Oct 3 '14 at 2:14
8
$\begingroup$

This simple and elegant solution relies on two results from Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$. and/or from Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$. both of which are derived from the well known Mittag-Leffler expansion of hyperbolic cotangent function (denote $\mathbb{W}=\mathbb{Z}/\{0\}$) :

$$\sum_{n\in\mathbb{W}}\frac{1}{n^2+z^2}=\frac{\pi\coth\pi z}{z}-\frac{1}{z^2}$$

valid for every complex $z$ exept the poles. The two later results from the links are in $\mathbb{W}$ formalism :

$$\begin{align}&\sum_{m\in\mathbb{W}}\frac{1}{(m-a)^2+b^2}=\frac{\pi\mathbin{\color{blue}{\sinh2\pi b}}}{b\left(\cosh2\pi b-\cos2\pi a\right)}-\frac{1}{a^2+b^2}\tag{$\varphi$}\\ \\ &\sum_{m\in\mathbb{W}} \frac{1}{m^4+c^4} = \frac{\pi}{\sqrt{2} \, c^3} \frac{\mathbin{\color{green}{\sinh\left(\sqrt{2} \pi c\right)+\sin\left(\sqrt{2} \pi c\right)}}}{\cosh\left(\sqrt{2} \pi c\right) -\cos\left(\sqrt{2} \pi c\right)}-\frac{1}{c^4}\tag{$\nu$}\end{align}$$

from these two formulae we are able by some simple manipulations obtain the sum $S$, for which we have:

$$S=\frac{1}{2}\sum_{n\in\mathbb{W}}\frac{1}{n^3}\frac{\mathbin{\color{red}{\sinh\left(\sqrt{2} \pi n\right)-\sin\left(\sqrt{2} \pi n\right)}}}{\cosh\left(\sqrt{2} \pi n\right) -\cos\left(\sqrt{2} \pi n\right)}$$

Note that

$$\mathbin{\color{red}{\sinh\left(\sqrt{2} \pi n\right)-\sin\left(\sqrt{2} \pi n\right)}}=2\mathbin{\color{blue}{\sinh\left(\sqrt{2} \pi n\right)}}-\left[\mathbin{\color{green}{\sinh\left(\sqrt{2} \pi n\right)+\sin\left(\sqrt{2} \pi n\right)}}\right]$$

i.e., this suggest using $2\varphi-\nu$. Choosing $a=b=n/\sqrt2$ in $\varphi$ and $c=n$ in $\nu$ we have for our sum:

$$\begin{align} S &=\frac{1}{2\pi\sqrt{2}}\sum_{n\in\mathbb{W}}\frac{2}{n^3}\left(\frac{1}{n}+\sum_{m\in\mathbb{W}}\frac{1}{(m-n/\sqrt2)^2+n^2/2}\right)-\frac{1}{n^3}\left(\frac{2}{n}+\sum_{m\in\mathbb{W}}\frac{2n^3}{m^4+n^4}\right)\\ \\ &=\frac{1}{\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2}\frac{1}{m^2-mn\sqrt{2}+n^2}-\frac{1}{m^4+n^4}\tag{1} \\ \\ &=\frac{1}{\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2}\frac{m^2+n^2}{m^4+n^4}-\frac{1}{m^4+n^4}\tag{2} \\ \\ &=\frac{1}{\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{m^2}{n^2}\frac{m^2}{m^4+n^4}\tag{3}\\ \\ &=\frac{1}{\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{1}{2}\left(\frac{m^2}{n^2}+\frac{n^2}{m^2}\right)\frac{1}{m^4+n^4}\tag{4}\\ &=\frac{1}{2\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{1}{m^2n^2}=\frac{1}{2\pi\sqrt{2}}\left(\sum_{n\in\mathbb{W}}\frac{1}{n^2}\right)^2=\frac{2}{\pi\sqrt{2}}\zeta^2(2)=\frac{\pi^2}{18\sqrt2}\tag{5} \end{align}$$

Explanations:

$(1)$ terms $2/n^4$ cancel each other; $(m-n/\sqrt2)^2+n^2/2=m^2-mn\sqrt{2}+n^2$

$(2)$ on the summation domain, lattice $(n,m)\in \mathbb{W}\times\mathbb{W}$ there is a symmetry $\sum_{n,m}=\sum_{-n,m}$, ergo by taking "average" $\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2}\frac{1}{m^2-mn\sqrt{2}+n^2}\!=\!\frac{1}{2}\!\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2}\left(\frac{1}{m^2-mn\sqrt{2}+n^2}+\frac{1}{m^2+mn\sqrt{2}+n^2}\right)$

$(3)$ uncomfortable terms $1/(n^4+m^4)$ cancels too

$(4)$ another symmetry = changing order of summation $n\longleftrightarrow m$

$(5)$ also $\zeta(2)=\sum_{n=1}^{\infty}1/n^2=\pi^2/6$

$\endgroup$

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.