Prove $\sum_{n=1}^\infty\frac{1}{(2n-1)^4}=\frac{\pi^4}{96}$

Question

Prove using Parseval identity applied to the functions: $x\,,|x|, x^2$ the convergence of the sum:

$$\sum_{n=1}^\infty\frac{1}{(2n-1)^4}=\frac{\pi^4}{96}\tag1$$

My attempt:

The identity of Parseval is:

$$2a_0^2+\sum_{n=1}^\infty (a_n^2+b_n^2)=\frac{1}{L}\int_{-L}^{L}f^2(x)$$

where $a_0,a_n,b_n $ are coefficients of fourier series.

Here i'm a little stuck. Can someone help me?

Answer 1

let $f=x^2$ and evaluate all the fourier coefficients and the integral in parsevals identity. move things around and you will get the sum from $n=1$ to infinity of $1/n^4$, or zeta(4). Notice that your sum is zeta(4) with only odd indices. You can find this given zeta(4) if you realize that the even indices in zeta(4) are equal to $1/16*zeta(4)$. So your sum is $(1-1/16)*zeta(4)=15/16*zeta(4)=pi^4 /96$