Consider $f(x):=\lvert x\rvert, x\in [-\pi,\pi]$. Then the Fourier series is $$ f(x)=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos((2n-1)x)}{(2n-1)^2}. $$
Now my task is to write down the related Parseval equation.
The general Parseval equation is $$ \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)=\frac{1}{\pi}\int_{-\pi}^{\pi}\lvert f(x)\rvert^2\, dx, $$ so here it is $$ \frac{\pi^2}{2}+\frac{16}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\ dx. $$
Is that the hole task or what?!
You got
$$0=f(0)=\frac\pi2-\frac4\pi\sum_{n=1}^\infty\frac1{(2n-1)^2}$$
and as commented by Avitus you have a numerical value for the above series. Now check that
$$\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\implies\ldots$$