2
$\begingroup$

I am looking to show: $\sum_{n=1}^∞ \frac{\coth(nπ)}{n^3} = \frac{7π^3}{180}$

There is a hint earlier that you are supposed to be using the function $f(z)=\frac{\cot z\coth z}{z^3}$. I have calculated the residue at the pole of order 5 at $z=0$ as $-\frac{7}{45}$, but I am unsure how to calculate the other residues, so I can use the residue theorem.

I think there is a simple pole whenever $z=\frac{(2n+1)π}{2}$, as this is when $\cot z=0$ but I just don't know how to find the residue here. I'm presuming my residues will lead to the sum I'm wanting to find coming out in some form when I apply the residue theorem, but I'm just not sure how to get there.

Thanks so much for any help in advance.

$\endgroup$
  • $\begingroup$ Surely you mean poles at $z=n\pi $, when $\cot z=\infty $, don't you? $\endgroup$ – user May 6 at 19:36
1
$\begingroup$

Observe that the integrand has simple poles at the points $z=n\pi$ and $z=i n\pi$, with $n\in\mathbb Z$, $n\ne0$.

The residues at the poles are computed as : $$ \lim_{z\to n\pi}(z-n\pi)\frac{\cot z\coth z}{z^3}=\lim_{\zeta\to0}\zeta\frac{\cot (\zeta)\coth(\zeta+n\pi)}{(\zeta+n\pi)^3}=\frac{\coth(n\pi)}{(n\pi)^3}, $$ and $$ \lim_{z\to in\pi}(z-in\pi)\frac{\cot z\coth z}{z^3}=\lim_{\zeta\to0}\zeta\frac{\cot (\zeta+in\pi)\coth(\zeta)}{(\zeta+i n\pi)^3}=\frac{\coth(n\pi)}{(n\pi)^3}, $$ where we used $$\begin{align} &\lim_{x\to0}x\cot x=\lim_{x\to0}x\coth x=1,\\ &\cot(x+n\pi)=\cot(x),\\ &\coth(x+in\pi)=\coth(x),\\ &\cot(i x)=-i\coth(x). \end{align} $$

With this and a suitable choice of the integration contour you will obtain: $$\begin{align} 4\sum_{n=1}^\infty\frac{\coth(n\pi)}{(n\pi)^3}&=\sum_{n=1}^\infty\left[\operatorname{Res}(f,-n\pi)+\operatorname{Res}(f,n\pi)+\operatorname{Res}(f,-in\pi)+\operatorname{Res}(f,in\pi)\right]\\ &=-\operatorname{Res}(f,0)=\frac7{45}. \end{align}$$

$\endgroup$
0
$\begingroup$

In this answer, it is shown that $$ \pi\coth(\pi n)=\frac1n+2n\sum_{k=1}^\infty\frac1{n^2+k^2}\tag1 $$ Therefore, since $\zeta(2)=\frac{\pi^2}6$, $$ \begin{align} \sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^2\!\left(n^2+k^2\right)} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2}\left(\frac1{n^2}-\frac1{n^2+k^2}\right)\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2n^2}-\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2\!\left(n^2+k^2\right)}\\ &=\frac12\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2n^2}\\[3pt] &=\frac12\zeta(2)^2\\[6pt] &=\frac{\pi^4}{72}\tag2 \end{align} $$ Thus, since $\zeta(4)=\frac{\pi^4}{90}$ $$ \begin{align} \sum_{n=1}^\infty\frac{\coth(\pi n)}{n^3} &=\frac1\pi\sum_{n=1}^\infty\frac1{n^4}+\frac2\pi\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^2\!\left(n^2+k^2\right)}\\ &=\frac{\pi^3}{90}+\frac{\pi^3}{36}\\[3pt] &=\frac{7\pi^3}{180}\tag3 \end{align} $$

$\endgroup$

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.