Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series)

Let $f(x)=x^2$ for $x\in(-\pi,\pi)$. Computing the Fourier coefficients gives

$$a_n=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 e^{i n x} dx=\frac{2 \cos(\pi n)}{n^2}=2\frac{(-1)^n}{n^2}$$

for $n\in\mathbb{Z}$, $n\not=0$, and $a_0=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 dx=\frac{\pi^2}{3}$.

Therefore $|a_n|^2=\frac{4}{n^4}$ for $n\in\mathbb{Z}$, $n\not=0$ and $|a_0|^2=\frac{\pi^4}{9}$.

By Plancherel/Parseval's theorem,

$$\frac{\pi^4}{9}+8\sum_{n=1}^\infty \frac{1}{n^4}=\sum_{n=-\infty}^\infty |a_n|^2=\frac{1}{2\pi}\int_{-\pi}^\pi x^4 dx=\frac{\pi^4}{5}$$

Simplifying, this gives

$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{8}\left(\frac{1}{5}-\frac{1}{9}\right)=\frac{\pi^4}{90}$$

We have $f \in L^2 \left[ -\pi, \pi \right]$ then \begin{align} \dfrac{1}{2}{A_0^2} + \displaystyle \sum_{n=1}^{\infty}{\left( A_n^2 + B_n^2 \right)} = \dfrac{1}{\pi} \int_{-\pi}^ \pi {f^2(x)} dx. \end{align} Here $A_n$ and $B_n$ are Fourier coefficients of $f$.

Let $f(x) = x^2$ for $x \in \left( -\pi, \pi \right)$. Then $f \in L^2 \left[ -\pi, \pi \right]$ and $$f(x) \sim \dfrac{{\pi}^2}{3} + 4{\displaystyle \sum_{n =1}^{\infty}{ \dfrac{(-1)^{n}}{n^2}\cos nx}}.$$ Therefore $$\dfrac{1}{2}\left(\dfrac{2 \pi^2}{3}\right)^2 +\displaystyle\sum_{n=1}^{\infty} \left( \dfrac{4.(-1)^n}{n^2} \right)^2 = \dfrac{1}{\pi} \int_{-\pi}^\pi{x^4}dx $$

So $$ \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^4}= \dfrac{\pi^4}{90}.$$