# Calculate $\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^2}$ using Parseval

The exercise asks me to calculate $$\displaystyle{\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^2}}$$ using the Fourier series of $$$$f(x) = \begin{cases} 0 & -\pi < x < 0 \\ \sin(x) & 0 \leq x < \pi \\ \end{cases}$$$$

which, in an earlier problem, was found to be $$\displaystyle{\frac{1}{4i}e^{ix}-\frac{1}{4i}e^{-ix} + \sum_{n\in\mathbb{Z}}\frac{1}{\pi(1-4n^2)}e^{2inx}}$$, and Parseval's Theorem. It is not clear to me, how Parseval could help me rearrange this into an equation to solve for $$\displaystyle{\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^2}}$$. On the right side of Parseval's I get $$\displaystyle{\int_{-\pi}^{\pi}\sin^2(x) dx = \pi}.$$

Any help on how to get started on this Problem would be much appreciated.

The answer is straightly forward. In fact $$c_n=\frac1{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx=\frac1{2\pi}\int_{0}^\pi e^{-inx}\sin xdx=-\frac{1+e^{-in\pi}}{2\pi(n^2-1)}, n\neq\pm1.$$ So $$c_{\pm1}=-\frac i4,c_{2n-1}=0, n\ge 2,$$ and $$c_{2n}=-\frac{1}{\pi(4n^2-1)}.$$ Note $$\sum_{n=-\infty}^\infty |c_n|^2=|c_0|^2+2|c_1|^2+2\sum_{n=1}^\infty\frac1{\pi^2(4n^2-1)^2}=\frac{1}{\pi^2}+\frac18+2\sum_{n=1}^\infty\frac1{\pi^2(4n^2-1)^2}$$ $$\int_{-\pi}^\pi|f(x)|^2dx=\frac{\pi}{2}.$$ By Parseval's Identity, $$2\pi\sum_{n=-\infty}^\infty |c_n|^2=\int_{-\pi}^\pi|f(x)|^2dx$$ one has $$2\pi\bigg[\frac{1}{\pi^2}+\frac18+2\sum_{n=1}^\infty\frac1{\pi^2(4n^2-1)^2}\bigg]=\frac\pi2$$ from which one derives $$\sum_{n=1}^\infty\frac1{(4n^2-1)^2}=\frac{\pi^2-8}{16}.$$