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The following is a conjecture.

I would like to prove that

$$\sum_{n=0}^\infty \frac{1}{(2n+1)\operatorname{sinh}((2n+1)\pi)}=\frac{\log(2)}{8}.$$

Both sides agree to at least $100$ digits, so I suspect the identity is true. I have thought about expanding $1/\operatorname{sinh}$ as a series and switching the summation, but I can't get this to work.

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So you computed the lhs up to 100 digits and then guessed that it would be equal to rhs? – 1015 Feb 14 '13 at 23:00
@julien Yes. $ $ – Potato Feb 14 '13 at 23:01
Wow, I'll believe you then! – 1015 Feb 14 '13 at 23:01
Wolfram seems to agree with the conjecture. – Argon Feb 14 '13 at 23:30
Does Wolfram do it algebraically, or numerically? – 1015 Feb 14 '13 at 23:39
up vote 7 down vote accepted

One of the references in the paper Ethan mentions is:

"Some Infinite Series of Exponential and Hyperbolic Functions", I. J. Zucker, SIAM Journal on Mathematical Analysis 15 (1984), 406-413, DOI 10.1137/0515031.

In Table 1, this paper has a number of equations similar to the identity the questioner gives; in fact, this identity can be gotten from Table 1 by subtracting (T1.3) from (T1.4) and substituting $k=k'=1/\sqrt{2}$, $c=1$. Here, $k$, $k'$, $c$, and $q$ (to appear momentarily) are parameters of the lattice for a doubly periodic function and are related by $$ k^2+k'^2=1, \qquad c=\frac{K(k')}{K(k)},\qquad q=e^{-\pi c} $$ where $$ K(k)=\int_0^1 \frac{1}{\sqrt{(1-x^2)(1-k^2 x^2)}} \, dx $$ is the complete elliptic integral of the first kind.

The method Zucker uses to prove the identities in Table 1 is to take the logarithm of the product representation of a product of theta functions, expand the logarithm, and then rearrange the sum. To use this method to prove the questioner's identity, start from the following identity (derived from pp. 469, 479, Whittaker & Watson, 4th edition): $$ \frac{\theta_4(0,q)}{\theta_3(0,q)}=\sqrt{k'}=\prod_{n\ge 1} \frac{(1-q^{2n-1})^2}{(1+q^{2n-1})^2}. $$ Taking logarithms and expanding the logarithm as a power series gives \begin{eqnarray*} \log k'&=&\sum_{n, m\ge 1} \frac4m((-1)^m-1)q^{(2n-1)m}\\ &=&-\sum_{n\ge1,\ m\ge 0} \frac{8}{2m+1}q^{(2n-1)(2m+1)}\\ &=&-4\sum_{m\ge 0} \frac{1}{2m+1} \frac{2}{q^{-(2m+1)}-q^{2m+1}}. \end{eqnarray*} Set $k=k'=1/\sqrt{2}$; then $q=e^{-\pi}$. This gives the desired result.

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This is not a proof of your identity, but some resources you can use to probably prove it.

Re-writing the hyperbolic function in terms of exponentials and using some partial fraction decomposition, allows you to re-write that identity as, $$\sum_{n=0}^\infty\frac{1}{(2n+1)(e^{\pi(2n+1)}-1)}+\sum_{n=0}^\infty\frac{1}{(2n+1)(e^{\pi(2n+1)}+1)}=\frac{\ln(2)}{8}$$

Which should look alot like many Plouffe/Ramanujan identities,

If you have ever seen Ramanujan's identity, $$\zeta(3)=\frac{7\pi^3}{180}-2\sum_{n=1}^\infty\frac{1}{n^3(e^{2\pi n}-1)}$$ And similar identitys involving logarithm constants and lambert series' are given at the bottom of: (

For your particular identity if you use some of the identities proven here, (

You can probably prove your identity.

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Well, I attempted using residues, as I expressed here. Once it was pointed out to me my mistake, I went to work and set up another sum based on the infinite set of residues along the line $\Re{z} = -1/2$. It turns out that

$$\sum_{n=0}^\infty \frac{1}{(2n+1)\operatorname{sinh}((2n+1)\pi)}=\sum_{n=0}^\infty \frac{(-1)^n}{2 n} \tanh{\left ( \pi \frac{n}{2} \right )}$$

You can verify this in Mathematica or elsewhere. Unfortunately, I seemed to have swapped one tough sum for another. Sigh.

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