Not Your Average Summation

Relevant wiki: Digamma Function

$\begin{aligned} S & = \sum_{n=2}^\infty \frac 1{n^4-1} \\ & = \sum_{n=2}^\infty \frac 1{(n^2-1)(n^2+1)} \\ & = \frac 12 \sum_{n=2}^\infty \left(\frac 1{n^2-1} - \frac 1{n^2+1} \right) \\ & = \frac 12 \sum_{n=2}^\infty \frac 1{(n-1)(n+1)} - \frac 12 \sum_{n=2}^\infty \frac 1{(n-i)(n+i)} \\ & = \frac 14 \sum_{n=2}^\infty \left(\frac 1{n-1} - \frac 1{n+1}\right) - \frac 1{4i} \sum_{\color{#3D99F6}n=2}^\infty \left(\frac 1{n-i} - \frac 1{n+i} \right) \\ & = \frac 14 \left(1+\frac 12\right) - \frac 1{4i} \left({\color{#3D99F6}\sum_{\color{#D61F06}n=1}^\infty \frac 1{n-i}} - \frac 1{1-i} - {\color{#3D99F6}\sum_{\color{#D61F06}n=1}^\infty \frac 1{n+i}} + \frac 1{1+i} \right) & \small \color{#3D99F6} \text{Digamma function }\psi_0 (z+1) = \gamma - \sum_{n=1}^\infty \left(\frac 1{z+n} - \frac 1n\right) \\ & = \frac 38 + \frac 14 - \frac 1{4i} \left(\color{#3D99F6} \psi_0 (1+i) - \psi_0 (1-i) \right) & \small \color{#3D99F6} \psi_0 (1-z) - \psi_0 (z) = \pi \cot (\pi z) \\ & = \frac 58 + \frac 1{4i} \left(\color{#3D99F6} \frac 1i - \pi\cot (\pi i) \right) & \small \color{#3D99F6} \psi_0 (z+1) = \psi_0 (z) + \frac 1z \\ & = \frac 58 + \frac 14 + \frac 1{4i} \left( \frac {\pi i(e^{-\pi}+e^\pi)}{e^{-\pi}-e^\pi} \right) \\ & = \frac 78 - \frac \pi 4 \coth \pi \\ & \approx \boxed{0.0867} \end{aligned}$