I am practicing Fourier series and am trying to use the Fourier series for $f(x) = x^2$ to show that $\frac{\pi^2}{12} = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \dots$
What I have done so far is calculate that the fourier series for $x^2$ is
$$f(x) = \frac{\pi^2}{3} + 4\sum_{n=1}^\infty \frac{\cos(n\pi)}{n^2} \cos(nx)$$
And now in trying to represent $\frac{\pi^2}{12}$ here is what I have done.
I have noted that I probably shouldn't just plug in $x = \frac{\pi}{\sqrt{12}}$ because that will give weird answers for $\cos(nx)$ so I have tried to do stuff like
$$f(\frac{\pi}{2}) = \frac{\pi^2}{4} = \frac{\pi^2}{3} + 4(1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \dots)$$
so then I need to get rid of the $\frac{\pi^2}{3}$ and the 4 and I am trying to get to $\frac{\pi^2}{12}$. I could subtract out the $\frac{\pi^2}{3}$ to give,
$$-\frac{\pi^2}{12} = 4(1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \dots)$$
But it seems impossible to get rid of the 4 or to deal with the fact that I have negative $\frac{\pi^2}{12}$. Does anyone have any suggestions?
You haven't evaluated correctly at $x = \frac\pi2$. We have $$ \sum_{n=1}^\infty \frac{\cos(n\pi)}{n^2} \cos(nx) = \frac{\cos(\pi)}{1^2}\cos(\frac\pi2) + \frac{\cos(2\pi)}{2^2}\cos(\pi) + \cdots\\ = 0 -\frac14 + 0 + \frac1{16} +0 + \frac{1}{36} + \cdots $$ Multiply by $4$, and we get $$ -1+\frac14-\frac19+\cdots $$ which is what you want $-\frac{\pi^2}{12}$ to be.