Proof for $\sum_{k=2}^\infty \frac{1}{k^4-1}= \frac{7}{8}-\frac{\pi}{4}\coth(\pi)$

How can this identity be derived? I have been searching the internet but I have no clue where to find a proof for this identity. Any help is highly appreciated. $$\sum_{k=2}^\infty \frac{1}{k^4-1}= \frac{7}{8}-\frac{\pi}{4}\coth(\pi)$$

• Use partial fractions to break up the sum as $$\frac{1}{4}\sum_{k \geq 2} \left( \frac{1}{k-1} - \frac{1}{k+1} \right) = \frac{3}{8},$$ and $$\frac{1}{4i}\sum_{k \geq 2} \left(\frac{1}{k-i} - \frac{1}{k+i}\right).$$ This last one you can probably relate to the residues of $\coth(z)$... Have a look at : math.stackexchange.com/questions/208317/… – Dzoooks Jun 12 '18 at 18:59
• I'd expect complex analysis on the function $f(z) = \frac{\cot(\pi z)}{z^4 - 1}$ to be able to get the answer. – Daniel Schepler Jun 12 '18 at 19:42

$$\sum_{k\geq 2}\frac{1}{k^2-1} = \frac{1}{2}\sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\frac{3}{4}\tag{1}$$ is trivial by telescoping and $$\sum_{k\geq 0}\frac{1}{k^2+1} = \frac{\pi\coth \pi+1}{2}\tag{2}$$ is a straightforward consequence of the Poisson summation formula, since the Laplace distribution and the Cauchy distribution are conjugated via the Fourier transform.

• This was very helpful. It is possible to decompose fractions like 1/(k^8-1) into something like this, so the value of their limit can be given in close form? – Mister Set Jun 12 '18 at 19:21
• @MisterSet: the approach is exactly the same. – Jack D'Aurizio Jun 12 '18 at 19:23
• Isn' t (1) equal to 3/4? – Mister Set Jun 12 '18 at 19:37
• @MisterSet: any good book about Fourier series and the Fourier transform covers that topic pretty well. I can recommend Stein's book, for instance. Anyway, there are plenty of alternative approaches, based on Cauchy's integral formula or the Weierstrass product $$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right).$$ – Jack D'Aurizio Jun 12 '18 at 19:38

Using the residue theorem, we have

\begin{align} \oint_{|z|=N+1/2}\frac{\cot(\pi z)}{(z^2+1)}\,dz&=2\pi i \sum\text{Res}\left(\frac{\cot(\pi z)}{z^2+1}\right)\\\\ &=2\pi i \left(\frac{2\cot(\pi i)}{2i}+\sum_{n=-N}^N \frac{1}{\pi(n^2+1)}\right)\tag1 \end{align}

As $N\to \infty$, the integral on the left-hand side of $(1)$ vanishes. Hence, we have

$$\sum_{n=-\infty}^\infty \frac{1}{n^2+1}=\pi \coth(\pi)$$

which after exploiting symmetry yields

$$\sum_{k=2}^\infty \frac{1}{n^2+1}=\frac{\pi\coth(\pi)}{2}-1\tag 2$$

Finally, using $\frac{1}{k^4-1}=\frac12\left(\frac1{k^2-1}-\frac{1}{k^2+1}\right)$ along with $(2)$ and the value of the telescoping series $\sum_{k=2}^\infty \frac{1}{k^2-1}$ yields the coveted result.

• Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Jun 16 '18 at 1:00