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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.

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  • $\begingroup$ What is the argument of $\cosh$ in the denominator? $\endgroup$
    – user64494
    Oct 12, 2013 at 15:40
  • $\begingroup$ @user64494 sorry I have to change $\endgroup$ Oct 12, 2013 at 15:42
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    $\begingroup$ @user64494 I to admire the beauty of it..... $\endgroup$ Oct 13, 2013 at 2:55
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    $\begingroup$ I think it would be helpful to provide a source for this problem. $\endgroup$
    – Slade
    Sep 30, 2014 at 5:35
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    $\begingroup$ @Slade I got it playing around with mathematica and the Inverse symbolic calculator. $\endgroup$ Oct 1, 2014 at 6:45

2 Answers 2

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

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  • $\begingroup$ Did not see the connection to the Ramanujan problem coming! Very nice. $\endgroup$ Oct 1, 2014 at 12:12
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    $\begingroup$ @Semiclassical It was obvious - the OP is a huge fan of Ramanujan ;) $\endgroup$ Oct 1, 2014 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.) $\endgroup$ Oct 1, 2014 at 13:52
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    $\begingroup$ You elucidated the mystery by pointing towards an enigma and showing that the two are the same. $\endgroup$
    – Lucian
    Oct 1, 2014 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 :) $\endgroup$ Oct 3, 2014 at 2:14
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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 $\zeta(2)=\sum_{n=1}^{\infty}1/n^2=\pi^2/6$

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