Prove that $\sum_n 1/(2n − 1)^4 = \pi^4/96$

Question

Attempt: I found the Fourier series for $f(x) = \lvert x\rvert$

$a_0 = (1/\pi)\int_{-\pi}^{\pi} \lvert r\rvert\,dr = \pi$

$a_n = (1/\pi)\int_{-\pi}^{\pi} \lvert r\rvert\cos(nr)\,dr = (2/\pi n^2)((-1)^n – 1)$

$b_n = (1/\pi)\int_{-\pi}^{\pi} \lvert r\rvert\sin(nr)\,dr = 0$

$f(x) = \pi/2 + \sum_n (2/\pi n^2)((-1)^n – 1)\cos(nx)$

The professor said to use this with Parseval's identity to do the proof. How do I do this?

Answer 1

Note that $a_n = 0$ if $n$ is even. Now you basically directly apply the Parseval's identity to $f(x) = |x|$ and it should give you the result after one rearranging of the terms.