I'm having a few questions regarding the following problem:
Calculate the Fourier series of $f(t)=|t|$ in $[-\pi, \pi)$ and then prove with $$\sum_{k=-n}^n |ck^2| = \frac{1}{2\pi}\int_0^\pi{|f(x)|^2}\,\mathrm dx$$ that $$\sum_{k=0}^{+\infty} \frac{1}{(2k+1)^4}=\frac{\pi^4}{96}.$$
I have calculated the Fourier series: $$f(t)=\pi+\sum_{k=0}^n -\frac{4}{\pi k}\cos(kx)$$ but I don't see any connection to the second part of the problem.
What have I missed or done wrong?
$$\begin{align*}&c_n=\frac1\pi\int_{-\pi}^\pi |t|e^{-int}\,dt=\frac1\pi\int_0^\pi t\left(e^{int}+e^{-int}\right)dt=\frac2\pi\int_0^\pi t\cos nt\,dt\\{}\\ &=\overbrace{\left.\frac{2t}{\pi n}\sin nt\right|_0^\pi}^{=0}-\frac2{\pi n}\int_0^\pi\sin nt\,dt=\left.\frac2{\pi n^2}\cos nt\right|_0^\pi=\\{}\\&= \frac2{\pi n^2}\left[(-1)^n-1\right]=\begin{cases}\;\;\;\;\;0,&n\;\text{even}\\{}\\-\cfrac4{\pi n^2},&n\;\text{odd}\end{cases}\end{align*}$$
and finally (observe the limits of the integral! It is not zero in the lower limit as you can see here)
$$2\sum_{n=0}^\infty\frac{16}{\pi^2(2n+1)^4}=\color{red}{\sum_{-\infty}^\infty|a_n|^2=\frac1{2\pi}\int_{-\pi}^\pi t^2\,dt}=\frac1{6\pi}2\pi^3=\frac{\pi^2}3\implies$$
$$\sum_{n=0}^\infty\frac1{(2n+1)^4}=\frac{\pi^4}{96}$$