Using, $f(t) = \begin{cases} \pi^2 & -\pi<t<0 \\[2ex] (t-\pi)^2 & 0<t<\pi \end{cases}$
The value of $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$ needs to be determined.
I found the Fourier series as $$f(x)=\frac{2\pi^2}{3}+\sum_{i=1}^n \frac{2}{i^2}\cos(ix)+\sum_{i=1}^n b_i\sin(ix)$$
I tried setting the values $x=\pm\pi$, only to get different (also incorrect) answers in both cases.
Any help is appreciated.
Note: $b_i$ left out as it seemed irrelevant for the purposes of the question. Please let me know if it is otherwise.
One can just use $$f(x)=\frac{1}{2} x^2 (-\pi<x \leq \pi).$$ Then, the Fourier series will be $$f(x)=\frac{\pi^2}{6}+2 \sum_{k=1}^{\infty} \frac{(-1)^k}{k^2} \cos(kx).$$ When $x=0$, $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}.$$