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

Question

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

Answer 1

$$\sum_{k\geq 2}\frac{1}{k^2-1} = \frac{1}{2}\sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\frac{3}{4}\tag{1} $$ 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.

Answer 2

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.