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**Random Variable** I found a proof, but it's a bit convoluted.

The Fourier series of $\displaystyle |x| $ on $\displaystyle (-\pi, \pi] $ is $\displaystyle |x| = \frac{\pi}{2} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos (2n+1)x}{(2n+1)^{2}} $

now integrate both sides from 0 to x

$\displaystyle \frac{x^{2}}{2} = \frac{\pi}{2} x - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\sin (2n+1)x}{(2n+1)^{3}} $

so $\displaystyle \sum_{n=0}^{\infty} \frac{\sin (2n+1)x}{(2n+1)^{3}} = \frac{\pi x}{8}(\pi-x) $

now let $\displaystyle x = \frac{\pi}{2} $

then $\displaystyle \sum_{n=0}^{\infty} \frac{\sin (2n+1) \frac{\pi}{2} }{(2n+1)^{3}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{3}} = \frac{\pi^{3}}{32} $