Function odd and $2\pi$-periodic, Fourier series and values of the sum I calculated, for the function ($2\pi$-periodic and odd) $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=1$ if $x \in ]0,\pi[$ and $f(x)=0$ if $x=\pi$, the Fourrier serie which is $\dfrac{2}{\pi} \sum_{k \geq 1} \dfrac{1-(-1)^k}{k} \sin(kt) = \dfrac{4}{\pi} \sum_{p \geq 0} \dfrac{\sin(t(2p+1))}{2p+1}$. By Dirichlet theorem, I found $\sum_{k \geq 0} \dfrac{(-1)^k}{2k+1} = \dfrac{\pi}{4}$. But how to find $\sum_{k \geq 0} \dfrac{1}{(2k+1)^2}$ with the previous result ? Do you have food for thought ?

Parseval's theorem comes to mind. The sum of squares of the coefficients of a Fourier series is related to the integral of $f^2$ over the interval $[0, \pi]$. This can be seen directly by squaring the series and observing that the cross-products integrate to zero over $[-\pi, \pi]$ (the orthogonality of trigonometric functions).