Find the Fourier series of
$$f(x)=x(\pi-x) ,\text{ } -\pi \le x \le \pi$$
Sketch the graph of the series for $x \in [-3\pi, 3\pi]$.
By evaluating $f(x)$ at particular points, deduce the sum of the following series
(i) $$\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+...$$
(ii) $$\sum^{\infty}_{n=1}\frac{1}{n^2}$$
$$a_n=\frac{1}{\pi}\int^{\infty}_{-\infty}(x-\pi-x^2)\cos{nx}\text{ }dx$$
$$=\int^{\infty}_{-\infty}x\cos{nx}\text{ }dx-\frac{1}{\pi}\int^{\infty}_{-\infty}x^2\cos{nx}\text{ }dx$$
$$=-\frac{4}{n}(-1)^n$$
$$a_0= \frac{2}{\pi}\int^{\infty}_{-\infty}(x\pi-x^2)\text{ }dx=-\frac{4\pi^2}{3}$$
$$b_n=\int^{\infty}_{-\infty}x\cos{nx}\text{ }dx-\frac{1}{\pi}\int^{\infty}_{-\infty}x^2\cos{nx}\text{ }dx=-\frac{4}{n^2}(-1)^n$$
$$f(x)=-\frac{4\pi^2}{3}-4\sum^{\infty}_{n=1}\frac{(-1)^n}{n}\cos{nx}+\frac{(-1)^n}{n^2}\sin{nx}$$
The graph is ok for me.
Now, to prove the series
Putting $x=\frac{\pi}{2}$ (pt. of cty)
$$\frac{\pi}{2}=-\frac{4\pi^2}{3}-4\sum^{\infty}_{n=1}\frac{(-1)^n}{n^2}.(-1)^n$$
$$\sum^{\infty}_{n=1}\frac{1}{n^2}=-\frac{\pi}{24}(3+8\pi)$$
But I am unsure of this. And I need some hints for the other series.