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.
$$\sum_{k=2}^\infty  \frac{1}{k^4-1}= \frac{7}{8}-\frac{\pi}{4}\coth(\pi)$$
 A: $$\sum_{k\geq 2}\frac{1}{k^2-1} = \frac{1}{2}\sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\frac{3}{4}\tag{1} $$
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. 
A: 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.

