Prove that fourier serie $f(x)$ is $\pi^2 / 6 = 1 + 1/2^2 + 1/3^2 + \ldots$

Question

i have some difficults to prove that the fourier series of

$$ f(x) = \begin{cases} 0, & -\pi < x \leq 0, \\ x^2, & 0 < x \leq \pi. \end{cases} $$

can be aproximated to

$$ \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \ldots$$

My attempt:

I computed the fourier series of $f$ and i get:

$$f(x) = \frac{\pi^2}{6} + \sum_k^\infty \left(\frac{2(-1)^k}{k^2}\cos(kx) + \frac{1}{\pi}\left[\frac{-\pi^2k^2(-1)^k + 2(-1)^k-2}{k^3}\right]\sin(kx)\right)$$

If $x = 0$ then

$0 = f(0) = \frac{\pi^2}{6} + \displaystyle\sum_k^{\infty} \frac{2(-1)^k}{k^2}$

that implies

$\frac{\pi^2}{6} = -\displaystyle\sum_k^{\infty} \frac{2(-1)^k}{k^2}$

Here i'm stuck, can someone help me?

Answer 1

\begin{align} & \sum_{k=1}^{\infty} \dfrac{1}{k^2} = \dfrac{\pi^2}{6}. \tag{A}\label{A}\\ & \sum_{k=1}^{\infty} \dfrac{1}{(2k)^2} = \dfrac{\pi^2}{24}. \tag{B}\label{B}\\ & \text{Take } (\ref{A})-2(\ref{B}): \\ & \sum_{k=1}^{\infty} \left(\dfrac{1}{(2k-1)^2}+\dfrac{1}{(2k)^2}\right) - 2\sum_{k=1}^{\infty} \dfrac{1}{(2k)^2} = \sum_{k=1}^{\infty} \left(\dfrac{1}{(2k-1)^2} - \dfrac{1}{(2k)^2}\right) \\ & = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} = \dfrac{\pi^2}{6} - \dfrac{\pi^2}{12} = \dfrac{\pi^2}{12}. \\ \ \\ \therefore \; & \dfrac{\pi^2}{6} + 2\sum_{k=1}^{\infty} \dfrac{(-1)^k}{k^2} = 0. \blacksquare \end{align}