Evaluating Fourier Series

Question

By evaluating the Fourier series of $g(x)$ at $x = 0$ show that: $$\dfrac{\pi^2}{12} = \sum_{n=1}^\infty\dfrac{(-1)^n+1}{n^2}$$

I have that $g(x)=f'(x)$ where $$f(x) = \dfrac \pi2 - \dfrac4\pi\sum_{n=1}^\infty \dfrac{\cos(nx)}{n^2}$$

I'm quite confused as subbing in $x=0$ obviously gets $0$.

Answer 1

This is more of a comment than an actual answer

This is a graph of $f(x)$, so far I can tell $f(x)$ is not even differentiable at $x= 0$.

Answer 2

If $$g(x)=\frac12x(x-\pi)=a_0+\sum_{k=1}^{\infty}a_k\cos kx$$ For $0\le x\le\pi$, then $$\int_0^{\pi}\frac12x(x-\pi)\cdot1dx=\left[\frac16x^3-\frac{\pi}4x^2\right]_0^{\pi}=-\frac{\pi^3}{12}=\pi a_0$$ $$\begin{align}\int_0^{\pi}\frac12x(x-\pi)\cos kx\,dx&=\left[\left(\frac1{2k}x(x-\pi)-\frac1{k^3}\right)\sin kx+\frac{x-\frac{\pi}2}{k^2}\cos kx\right]_0^{\pi}\\ &=\frac{\pi}{2k^2}\left[1+(-1)^k\right]=\frac{\pi}2a_k\end{align}$$ So $$g(x)=-\frac{\pi^2}{12}+\sum_{k=1}^{\infty}\frac{1+(-1)^k}{k^2}\cos kx$$ And we have $$g(0)=0=-\frac{\pi^2}{12}+\sum_{k=1}^{\infty}\frac{1+(-1)^k}{k^2}$$ So I think that expression for $f(x)$ was up there is wrong somehow -- there never was an assertion about the function the Fourier series was intended to represent was.