After reworking the problem many times I keep getting the same (incorrect?) answer.
So the problem as stated is
Find the Fourier expansion of : $$ f(x) = \begin{cases} x &\text{ if }0 < x < \pi,\\ 2(\pi - x) &\text{ if }\pi < x < 2\pi \end{cases} $$
I get the following Fourier coefficients: \begin{align} a_0 &= \frac{-π}{4} \\ a_n &= \frac{ 1 - (-1)^n}{π n^2} \\ b_n &= \frac{4 + 3(-1)^n}{n} \end{align}
They are apparently wrong since I'm meant to show that $$ \frac{π^2}{8} = \sum_{n=0}^{+\infty} \frac{1}{(2n-1)^2} $$
I've checked the jump discontinuities $ \frac{1}{2(f^+(\pi) + f^-(\pi))}$ and $\frac{1}{2(f^+(2\pi) + f^-(2\pi))}$ so it can only be an error in my coefficients?
Consider the shifted variant $f(x)=\pi-|x|$ on $[-π,π]$ as fundamental period.
Then, since it is an even function, $b_n=0$ and \begin{align} a_0&=\frac1{\pi}\int_0^\pi(\pi-x)dx\\&=\frac1{2\pi}[-(\pi-x)^2|_0^\pi=\frac{\pi}2 \\ a_n&=\frac{2}{\pi}\int_0^\pi(\pi-x)\cos(nx)dx \\&=\frac2{n\pi}[(\pi-x)\sin(nx)]_0^\pi+\frac2{n\pi}\int_0^\pi \sin(nx)\\ &=\frac2{n^2\pi}[-\cos(nx)]_0^\pi=\frac{2(1-(-1)^n)}{n^2\pi} \end{align} which implies $a_{2n}=0$ and $a_{2n+1}=\frac4{(2n+1)^2\pi}$
Comparing function and series at $x=0$ gives $$ \pi=f(0)=\frac\pi2+\frac4\pi\sum_{n=0}^\infty\frac1{(2n+1)^2} $$ which rearranged leads to $$ \frac{\pi^2}{8}=\sum_{n=0}^\infty\frac1{(2n+1)^2} $$