The result* is the fourier series for $f(x)=x+\pi$ on $(-\pi,\pi)$ is $\pi+\sum_{n=1}^{\infty} \frac{-2(-1)^n}{n} sin(nx)$
I know Parseval's identity derived from the fourier series of a function is $\frac{1}{L}\int_{-L}^{L}[f(x)]^2\,dx=\frac{(a_o)^2}{2}+\sum_{n=1}^{\infty}((a_n)^2+(b_n)^2)$
This is the extent of my knowledge and what I am given. I am unsure where to go from here, any help or nudge is greatly appreciated! If too little info, please let me know and I can try to give more.
Best Answer
Ok, so $a_0=\pi$, $a_n=0,$ $b_n=$$ \frac{-2(-1)^n}{n} $
So $a_0^2=\pi^2, b_n^2= \frac{4}{n^2}$
So let $f(x)=x+\pi,$ so $ f(x)^2= x^2+2\pi x+\pi^2$
Integrating $f(x)^2$ between $-\pi $ to $ \pi$ gives $\frac{8}{3}\pi^3 $ and so the LHS of the equality becomes $ \frac{8}{3}\pi^2$
So $ \frac {8}{3}\pi^2= \pi^2 +\sum_{n=1}^{\infty}\frac{4}{n^2} $
so $ \frac{2}{3}\pi^2$= $\sum_{n=1}^{\infty}\frac{4}{n^2}$
Divide both sides by 4 to get :
$\sum_{n=1}^{\infty}\frac{1}{n^2}= \frac{\pi^2}{6} $
As required.