<s>[INST] Fourier series expansion of $\frac{\pi^4}{96}$ and $\frac{\pi^4}{90}$ I have to prove that:
\begin{equation}
 \frac{\pi^4}{96}=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^4} \qquad \mbox{and} \qquad \frac{\pi^4}{90}=\sum_{n=1}^{+\infty}\frac{1}{n^4}
\end{equation}
My idea:
I've expanded $f(x)=|x|$ in a real Fourier series $\forall x\in Q:=[-\pi,\pi)$:
\begin{equation}
 f(x)=|x|=\frac{\pi}{2}-\sum_{n=0}^{+\infty} \frac{1}{(2n+1)^2\pi}\cos(nx)
\end{equation}
Now, using Parseval's Identity, I get:
\begin{equation}
 \frac{2\pi^2}{3}=\sum_{n=0}^{+\infty}\frac{16}{n^4\pi^2}+\frac{\pi^2}{2}
\end{equation}
hence:
\begin{equation}
 \frac{\pi^4}{96}=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^4}
\end{equation}
Now, how can I prove the second claim?
 [/INST] From what you've already proved:
$$\sum_{n=1}^\infty\frac 1{n^4}=\sum_{n=0}^\infty\frac 1{(2n+1)^4}=\sum_{n=1}^\infty\frac 1{(2n)^4},$$
whence 
$$\sum_{n=1}^\infty\frac 1{n^4}-\sum_{n=1}^\infty\frac 1{(2n)^4}=\frac{15}{16}\sum_{n=1}^\infty\frac 1{n^4}=\frac{\pi^4}{96}.$$
</s>