Exercise: Apply Parseval's identity to \(\sin(x/2)\) on \([-\pi, \pi]\) to compute an exact value for the infinite series \(\sum_{n = 1}^\infty n^2/(4n^2 - 1)^2\).


Proof. Parseval's Relation states that \begin{equation} ||f||^2 := \int_{-\pi}^\pi |f(x)|^2 dx = 2\pi \sum_{n = -\infty}^\infty|c_n|^2, \notag \end{equation} where \begin{equation} c_n = \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-inx} dx. \notag \end{equation} Letting \(f(x) = \sin (x/2)\), we then get that \begin{equation} c_n = \frac{1}{2\pi} \int_{-\pi}^\pi \sin(x/2) e^{-inx} dx = \frac{4in\cos(\pi n)}{\pi(4n^2 - 1)} \notag \end{equation} and \begin{equation} \int_{-\pi}^\pi |f(x)|^2 dx = 4, \notag \end{equation} so that \begin{equation} \sum_{n = -\infty}^\infty \left|\frac{4in\cos(\pi n)}{\pi(4n^2-1)}\right|^2 = \frac{2}{\pi}. \notag \end{equation} Because \(|g(n)|^2 = g^2(n)\) for any function \(g\), note that the above equation can be rewritten as \begin{align} \sum_{n = -\infty}^\infty \frac{(4in\cos(\pi n))^2}{(\pi(4n^2-1)^2} &= \sum_{n = -\infty}^\infty \frac{-4^2n^2\cos^2(\pi n)}{\pi^2(4n^2-1)^2} = \sum_{n = -\infty}^\infty \frac{-16 \cos^2(\pi n)}{\pi^2} \cdot \frac{n^2}{(4n^2-1)^2} = \frac{2}{\pi}. \notag \end{align} As \(\cos^2(\pi n) = 1\) for every \(n\), rewrite the above as \begin{equation} \sum_{n = -\infty}^\infty \frac{n^2}{(4n^2-1)^2} = \frac{\pi}{8}. \notag \end{equation}