Boundary Values & Fourier Series · Challenge problem

Find the Fourier series of $f(x) = x\sin x$ on $[-\pi, \pi]$ and evaluate it at $x = 0$ to find $\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{n^2 - 1} = \dfrac13 - \dfrac18 + \dfrac1{15} - \cdots$.