I need help with this problem. I'm asked if it is possible to find $\sum_{n=0}^\infty \frac{1}{n^2}$ with the Fourier series of $f(x)=1-|x|$ if $|x|\leq 1$ and $f(x+2)=f(x)$. I tried to do the Fourier series, but I didn't ebd up with the correct result.
Fourier series definition: $f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos\left(\frac{n\pi x}{L}\right)+\sum_{n=1}^\infty bn\sin\left(\frac{n\pi x}{L}\right)$
I found that:
$a_0=\frac{1}{L}\int_{-L}^{L}f(x)dx=\int_{-1}^1(1-|x|)dx=1 $ since $|x|=x $ in $[0,L]$.
$a_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos(\frac{n\pi x}{L}) dx=\int_{-1}^1(1-|x|)\cos(\frac{n\pi x}{1})dx=\frac{2-2\cos\left({\pi}n\right)}{{\pi}^2n^2}=\frac{2-2(-1)^n}{{\pi}^2n^2}$
$b_n=\frac{1}{L}\int_{-L}^{L}f(x)\sin(\frac{n\pi x}{L}) dx=\int_{-1}^1(1-|x|)\sin(\frac{n\pi x}{1})dx=0$
I know that $\sum_{n=0}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$, so I guess that the answer is no. Am I correct or what am I doing wrong?