Calculate Fourier series sum using Parseval's theorem

Question

I've used Parseval's theorem before, to calculate $\sum_{n=1}^\infty \frac{1}{n^4}$. Now I have to calculate $ \sum_{n=1}^\infty \frac{1}{(2n-1)^4}$ and I'm lost. I have to start with $f(x)=x, x\in[0,\pi]$. My Fourier coefficients are: $$a_0=\frac{\pi}{2}$$ $$a_n=\frac{(-1)^n-1}{\pi n^2}$$ $$b_n=\frac{(-1)^{n+1}}{n}$$ I've also managed to calculate $\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi ^2}{8}$. I've tried to do it: calculating $|a_0|^2$ is easy, but I have problem with $\sum_{n=1}^\infty |a_n|^2+|b_n|^2$, I get $\frac{-2}{\pi} \sum_{n=1}^\infty \frac{(-1)^n -1 -\pi}{n^2}$, which leads me nowhere. Am I doing something wrong? Or is it correct? In both cases, I think I need just a little help with this.

Answer 1

By comparing even numbers to odd numbers, we find that

\begin{align} \sum_{n \in \mathbb{N}} \frac 1 {n^4} &= \sum_{n \text{ even}} \frac 1 {n^4} + \sum_{n \text{ odd}} \frac 1 {n^4} \\ &= \frac 1 {2^4} \sum_{k \in \mathbb{N}} \frac 1 {k^4} + \sum_{k \in \mathbb{N}} \frac{1}{(2k - 1)^4} \end{align}

so that

$$\sum_{k \in \mathbb{N}} \frac 1 {(2k - 1)^4} = \frac {15}{16} \cdot \frac{\pi^4}{90} = \frac {\pi^4}{96}.$$