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


I have , somehow, to find the sum of $\sum_{n=1}^\infty \frac{1}{n^4}$ using Parseval's theorem.

I tried some things that didn't work so I won't post them.

Can you please explain me how do I find the sum of this series using Parseval's identity?

Thanks
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

$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{8}\left(\frac{1}{5}-\frac{1}{9}\right)=\frac{\pi^4}{90}$$
The series $\sum_{n\ge 1}\frac{1}{n^4}$ also appears in the following funny way. Consider the operator ${\rm d}^2/{\rm d}x^2$ on $[0,\pi]$ s.t. Dirichlet boundary conditions. Find Green's function $G(x,s)$. Find the normalized in $L^2[0.\pi]$ eigenfunctions $u_n(x)$ of the operator; the eigenvalues are $\lambda_n=n^2$.
Evaluate the integral $\int_0^\pi\int_0^\pi G^2(x,s){\rm d}x~{\rm d}s$ 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 $\sum_{n\ge 1}\frac{1}{n^4}=\frac{\pi^4}{90}$.