Closed Form series :$\sum_{n\ge 1} \frac{1}{n^4} $
let :$f(x)=x^2 $function periodic $2\pi$
we have serie fourier : $$f(x)=a_0 + \sum_{n\ge 1} \left({a_n\cos(nx) + b_n\sin(nx)}\right)$$
$$a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2 dx = \frac{\pi^2}{3}$$ $$a_n=\frac{1}{\pi}\int^{\pi}_{-\pi} x^2\cos(nx) dx = \frac{4(-1)^n}{n^2}$$
$$b_n=0$$ therfore :
$$f(x) = \frac{\pi^2}{3}+\sum_{n\ge 1} \frac{4(-1)^n}{n^2}\cos(nx)$$
using parseval
$$\frac{1}{\pi}\int^{2\pi}_{0} |f(x)| ^2 =\frac{1}{2}a^2_0 + \sum \left({a_n^2 + b_n^2}\right)$$
$$\frac{1}{\pi} \int_{0}^{2\pi} x^4 dx = \frac{\pi^4}{18} + \sum \frac{16}{n^4}$$
therfore : $$\sum \frac{16}{n^4} =\frac{32\pi^4}{5} - \frac{\pi^4}{18} $$
finnaly : $$\sum \frac{1}{n^4} = \frac{571\pi^4}{1440} $$ but using Wolfram aplha : $$\sum \frac{1}{n^4} = \frac{\pi^4}{90}$$
What is Error ?? any other methode prov it
Coming from a harmonic analysis background, I am really only comfortable with one particular normalization of the Fourier transform and of the Fourier series. For me, if $f\in L^2([0,1])$, then $\hat{f}(\omega)=\int_0^1 f(x)e^{-2\pi i \omega x} dx$.
So let us define the function $f:[0,1]\to\mathbb{R}$ by $f(x)=x^2$. Clearly, this is square-integrable and $\|f\|_2^2=\int_0^1 x^4 dx=\tfrac{1}{5}$.
Its Fourier series is given by $\hat{f}(n)=\int_0^1 x^2 e^{-2\pi inx}=\int_0^1 x^2\cos(2\pi nx)dx - i\int_0^1 x^2\sin(2\pi nx)dx$. For $n\in\mathbb{Z}\setminus\{0\}$, one can use integration by parts twice to show that $\int_0^1 x^2\cos(2\pi n x)dx=\tfrac{1}{2\pi^2 n^2}$ and that $\int_0^1 x^2\sin(2\pi n x) dx= -\tfrac{1}{2\pi n}$.
Therefore, $|\hat{f}(n)|^2=\tfrac{1}{4\pi^2 n^2}+\tfrac{1}{4\pi^4 n^4}$, for all $n\in \mathbb{Z}\setminus\{0\}$. Trivially, $\hat{f}(0)=\int_0^1 x^2 dx=\tfrac{1}{3}$, so $|\hat{f}(0)|^2 =\tfrac{1}{9}$.
Now Parseval's identity says that $\|f\|_2^2=\sum_{n\in\mathbb{Z}} |\hat{f}(n)|^2$. Note that in our case $|\hat{f}(-n)|^2=|\hat{f}(n)|^2$, so plugging everything in we get $\tfrac{1}{5}=\tfrac{1}{9}+\tfrac{1}{2\pi^2}\sum_{n=1}^\infty \tfrac{1}{n^2} + \tfrac{1}{2\pi^4} \sum_{n=1}^\infty \tfrac{1}{n^4}$. Using the fact that $\sum_{n=1}^\infty \tfrac{1}{n^2}=\tfrac{\pi^2}{6}$, we can rearrange to find $\sum_{n=1}^\infty \tfrac{1}{n^4} =2\pi^4\big(\tfrac{1}{5}-\tfrac{1}{9}-\tfrac{1}{12}\big)=\tfrac{\pi^4}{90}$.