Fourier series expansion of $\frac{\pi^4}{96}$ and $\frac{\pi^4}{90}$

Question

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?

Answer 1

how can I prove the second claim?

If you already know that $$ \frac{\pi^4}{90}=\sum_{n=1}^{+\infty}\frac{1}{n^4} $$ then observe that (by separating even and odd integers) $$ \sum_{n=1}^{+\infty}\frac{1}{n^4}=\sum_{n=1}^{+\infty}\frac{1}{(2k)^4}+\sum_{n=0}^{+\infty}\frac{1}{(2k+1)^4}. $$ Can you take it from here?

Answer 2

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 $$\sum_{n=1}^\infty\frac 1{n^4}-\sum_{n=1}^\infty\frac 1{(2n)^4}=\frac{15}{16}\sum_{n=1}^\infty\frac 1{n^4}=\frac{\pi^4}{96}.$$