1
$\begingroup$

Is there a closed form for this sum?

$$\sum _{n=1}^{\infty }\frac{\coth (xn)}{n^3}$$

I have tried using

$$x\coth \left(xn\right)=\frac{1}{n}+2n\sum _{k=1}^{\infty }\frac{1}{n^2+\left(\frac{\pi }{x}\right)^2k^2}$$

To solve but got stuck at evaluating

$$\sum _{n=1}^{ \infty}\sum _{k=1}^{\infty }\frac{1}{k^2\left(n^2+\left(\frac{\pi }{x}\right)^2k^2\right)}$$

$\endgroup$
2
  • $\begingroup$ @DougM: Assuming $x\neq0$, $\coth{nx}\to\pm1$ as $n\to\infty$. $\endgroup$ – John Bentin Aug 4 '16 at 16:31
  • $\begingroup$ Sorry, I had $\cosh x$ in my mind. $\endgroup$ – Doug M Aug 4 '16 at 16:37
5
$\begingroup$

We may consider that our series equals $$ S(x)=\frac{d}{dx}\sum_{n\geq 1}\frac{\log(\sinh(nx))}{n^4}=\frac{\zeta(4)}{x}+\frac{d}{dx}\sum_{n\geq 1}\frac{1}{n^4}\sum_{m\geq 1}\log\left(1+\frac{n^2 x^2}{\pi^2 m^2}\right) \tag{1}$$ where in the second step we exploited the Weierstrass product for $\frac{\sinh(z)}{z}$. We get: $$\begin{eqnarray*} S(x)&=&\frac{\zeta(4)}{x}+2x\sum_{n,m\geq 1}\frac{1}{m^2 n^2\pi^2+n^4 x^2}\\&=&\frac{\zeta(4)}{x}+2x\left(\frac{5\pi^2+x^2}{180}-\frac{x}{2\pi^2}\sum_{m\geq 1}\frac{1}{m^3}\,\coth\left(\frac{m\pi^2}{x}\right)\right)\tag{2}\end{eqnarray*} $$ and the same can be achieved by applying the Poisson summation formula, but I am not expecting a great simplification in the last sum (on $m\geq 1$) unless $x$ is a particular integer multiple of $\pi$.
However, it is interesting to notice that the above formula leads to: $$ S(\pi) = \sum_{n\geq 1}\frac{\coth(\pi n)}{n^3}=\color{red}{\frac{7\pi^3}{180}},\tag{3} $$ for instance.

$\endgroup$
2
  • 1
    $\begingroup$ why on earth people didn't upvote more on such nice answers...(+1) $\endgroup$ – tired Aug 4 '16 at 23:35
  • $\begingroup$ I up vote because I got a question on the general of this question. $\endgroup$ – gymbvghjkgkjkhgfkl Aug 5 '16 at 22:37

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.