[Math] Find the sum of a series using Fourier series

Question

Calculate the Fourier series of the function $f(x)=\sin{(ax)}$ on the interval $[-\pi,\pi]$ and then calculate:
$$\sum_{n=1}^{\infty}{\frac{n^{2}}{(4n^2-1)^2}}$$

I've already calculated the Fourier series for this function:

$$\sin{(ax)}=\frac{1}{\pi}\sum_{n=1}^{\infty}\frac{2n \sin{(a \pi)} (-1)^{n}}{a^{2}-n^{2}}\sin{(nx)}$$

I can't really find a suitable $a$ to sum the series in the problem. For one thing, the polynomial in the denominator of the series has a higher degree than in the Fourier series. I've tried using partial fractions but the degrees still differ.

Answer 1

Solution: Use Parseval's identity.

Set $f(x)=\sin\left(ax\right)$ with $a=\dfrac 1 2$, for all $x\in [-\pi, \pi]$. Find $b_n=\dfrac{(-1)^{n+1}8n}{\pi(4n^2-1)}$ for all $n\in \mathbb N$ and $\displaystyle \dfrac 2 \pi\int \limits_0^\pi\left(\sin\left(\dfrac x 2\right)\right)^2\mathrm dx=1$.