Prove that $\sum_{k=1}^\infty\frac{1}{16k^4 - 1} = \frac{1}{2} - \frac{\pi}{8}\coth(\frac{\pi}{2})$

Question

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.

Answer 1

You have used Parseval's theorem incorrectly. Try another approach: find the Fourier series for $f(x)=\cos \frac{x}{4}$, $-\pi\leq x\leq\pi$ and set there $x=\pi$.

Answer 2

For $\xi\in[0,1]$ and $k\in\Bbb R$, we have \begin{equation*} \frac{1}{k^4}+\sum_{n=1}^{\infty}\frac{2(-1)^{n+1}\cos(n\pi\xi)}{\left(n\pi\right)^4-k^4}=\frac{\sinh k\cos(k\xi)+\sin k\cosh(k\xi)}{2k^3\sinh k\sin k} \end{equation*} The above is just the Fourier cosine series for the expression on the right hand side (where the coefficients are obtained by using the usual orthogonality relations). By substituting $\xi=1$ and $k=\pi/2$ in the above relation, we get the desired result.

Answer 3

Another way...

$$8S=\sum_{k=1}^\infty\frac{8}{16k^4 - 1} \\ = \sum_{k=1}^\infty\frac{1}{(\tfrac i2)^2-k^2} -\sum_{k=1}^\infty\frac{1}{(\tfrac 12)^2-k^2} \\=\frac\pi{2\tfrac i2}\cot(\pi\tfrac i2)-\frac 1{2(\tfrac i2)^2}-\left(\frac\pi{2\tfrac 12}\cot(\pi\tfrac 12)-\frac 1{2(\tfrac 12)^2}\right) \\=4 -\pi\coth\left(\frac{\pi}{2}\right)$$

By using the partial fraction decomposition $$\pi x\cot\pi x=1+\sum_{k=1}^\infty\frac{2x^2}{x^2-k^2}$$