I have that,
$$f(x) = 1-(\frac{x^2}{\pi^2})$$ where it has period $2L = 2\pi$
I found that my Fourier series for $f(x)$ is $$\frac{2}{3} + \sum_{n = 1}^{\infty}\frac{-4(-1)^n}{\pi^2 n^2}\cos(nx)$$ (1)
I am to find that,
$$\frac{\pi^2}{12} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$$ (2)
I have no idea how to proceed from this point onwards,
Trying parseval's identity yields me:
$$\frac{28}{48} = \sum_{n=1}^{\infty}\frac{16(-1)^{2n}}{\pi^4 n^4}$$ (3)
Which unless i made a mistake doesnt yield anything obvious to me. Could someone point me in the right direction?