I have seen this post Fourier series expansion of $\frac{\pi^4}{96}$ and $\frac{\pi^4}{90}$ but it seems to skip some steps that I don't understand. I also looked at some others, but I haven't found one that exactly answers what I need.
Using $f(e^{i\theta}) = |\theta|$. I found the Fourier series to be
$$|\theta| = \frac{\pi}{2} – \frac{2}{\pi} \sum_{k = -\infty}^{\infty} \frac{e^{ik\theta}}{k^2}$$
$c_k = \begin{cases}
0 & \ \mathrm{if} \ k \ \mathrm{is \ even} \\
\frac{-2}{\pi k^2} & \ \mathrm{if} \ k \ \mathrm{ is \ odd} \\ \end{cases}$
So, one could say
$$|\theta| = \frac{\pi}{2} – \frac{2}{\pi} \sum_{k = -\infty}^{\infty} \frac{e^{ik\theta}}{(2k+1)^2}$$
Now, for Parseval's identity, we need $|c_k|^2 = \frac{4}{\pi^2 (2k+1)^4}$ and $\frac{1}{2\pi} \int_{-\pi}^{\pi} |f(e^{i\theta})|^2 d\theta = \frac{\pi^2}{3}$.
So,this means $$\frac{\pi^2}{3} = \sum_{k = -\infty}^{\infty} \frac{4}{\pi^2(2k+1)^4}.$$
So,
$$\frac{\pi^4}{12} = \sum_{k = -\infty}^{\infty} \frac{1}{(2k+1)^4}.$$
From here is where I am having problems. I'm unsure of how to convert this Fourier series to its equivalent starting at $1$. I thought it was just
$$\frac{\pi^4}{12} = 2\sum_{k = 1}^{\infty} \frac{1}{(2k+1)^4},$$ but this is of course not true as the result would be
$$\frac{\pi^4}{24} = \sum_{k = 1}^{\infty} \frac{1}{(2k+1)^4},$$ which is not the correct answer. (Edit: This part is correct, the answer is just not coming out correctly somehow.) I'm really not sure what the issue is. I did the same process using $f(e^{i\theta}) = \begin{cases}
-1 & \ -\pi < \theta < 0 \\
1 & \ 0 < \theta < \pi \end{cases}$ to prove the series for $\frac{\pi^2}{8}$, but it does not seem to be working here. Does anyone see what's happening?
Best Answer
To summarize what happened in the comments, the problem is that you forgot to factor in the $c_0^2$ term when doing your series.
$\pi^2/3=\pi^2/4+2\sum_{k=1}^{\infty}c_k^2$
here we noted that $c_{-k}=c_k$ so we grouped the sum together to get a series from $1$ to $\infty$.
Now, $$\frac{\pi^2}{3}-\frac{\pi^2}{4}=2\sum_{n=1}^{\infty} \frac{4}{\pi^2(2n+1)^4}$$ (here $k=2n+1$ as we omit all the even terms).
So we get $$\frac{\pi^4}{96}=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^4}$$