Evaluate ∑n=1∞1n4 using Parseval's theorem (Fourier series).

We have ${\displaystyle f\in L^2[-\pi ,\pi ]}$ then
${\displaystyle \frac{1}{2}{A}_0^2+\sum _{n=1}^{\mathrm{\infty }}(A_n^2+B_n^2)=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}^2\left(x\right)dx}$
Here ${\displaystyle A_n}$ and ${\displaystyle B_n}$ are Fourier coefficients of f.
Let ${\displaystyle f\left(x\right)=x^2}$ for ${\displaystyle x\in (-\pi ,\pi )}$. Then ${\displaystyle f\in L^2[-\pi ,\pi ]}$ and
${\displaystyle f\left(x\right)\sim \frac{{\pi }^2}{3}+4\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}\cos nx}$
Therefore
${\displaystyle \frac{1}{2}{\left(\frac{2{\pi }^2}{3}\right)}^2+\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{4{(-1)}^n}{n^2}\right)}^2=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^{4}dx}$
So
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4}=\frac{{\pi }^4}{90}}$

The series $\sum _{n\ge 1}\frac{1}{n^4}$ also appears in the following funny way. Consider the operator $d^2/d{x}^2$ on $[0,\pi ]$ s.t. Dirichlet boundary conditions. Find Green's function $G(x,s)$. Find the normalized in $L^2[0.\pi ]$ eigen functions $u_n(x)$ of the operator; the eigenvalues are ${\lambda }_n=n^2$.Evaluate the integral ${\int }_0^{\pi }{\int }_0^{\pi }{G}^2(x,s)dxds$ in two ways, a) using the explicit formula for $G(x,s)$ and b) using the representation $G(x,s)=\sum _{n\ge 1}\frac{u_n(x)u_n(s)}{{\lambda }_n}$.After simplifications, you will find $\sum _{n\ge 1}\frac{1}{n^4}=\frac{{\pi }^4}{90}$.