I want to prove that: $$\sum_{k=1}^\infty\frac{1}{16k^4 - 1} = \frac{1}{2} - \frac{\pi}{8}\coth\left(\frac{\pi}{2}\right)$$ Using the fourier series: $$\phi(x) = \begin{cases}0 & \text{if }-\pi<x<0, \\ \sin(x) & \text{if }0<x<\pi. \end{cases}$$ $$\phi(x) = \frac{1}{\pi} + \frac{1}{2} \sin(x)+ \frac {2}{\pi}\left( \sum_{n=1}^\infty\frac{\cos(2n x)}{4n^2 - 1}\right)$$ See: Fourier series of function $f(x)=0$ if $-\pi<x<0$ and $f(x)=\sin(x)$ if $0<x<\pi$
So far I have: $$\phi(\frac{\pi}{2}) = \sin\left(\frac{\pi}{2}\right) = 1$$ $$=\frac{1}{\pi} + \frac{1}{2}+ \frac {2}{\pi}\left( \sum_{n=1}^\infty\frac{\cos(\pi n)}{4n^2 - 1}\right)$$ $$ = \frac{1}{\pi} + \frac{1}{2}+ \frac {2}{\pi}\left( \sum_{n=1}^\infty\frac{(-1)^n}{4n^2 - 1}\right)$$ Then, using Parsaval's Theorem: $$\frac{1}{2\pi}\int_{-\pi}^{\pi}|\sin\left(\frac{\pi}{2}\right)| =\frac{1}{\pi} + \frac{1}{2}+ \frac {2}{\pi}\left( \sum_{n=1}^\infty|\frac{(-1)^n}{4n^2 - 1}|\right)$$ $$=\frac{1}{\pi} + \frac{1}{2}+ \frac {2}{\pi}\left( \sum_{n=1}^\infty\frac{1}{|4n^2 - 1|}\right)$$ I ma stuck here.
