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

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?

## closed as off-topic by Travis, B. Mehta, RRL, ancientmathematician, A. PongráczDec 31 '18 at 8:23

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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 $$(1-1/16)*zeta(4)=15/16*zeta(4)=pi^4 /96$$
• $\LaTeX \text{ Tip}:$ Use \zeta, \pi and \frac {a}{b} to obtain $\zeta$, $\pi$ and $\frac ab$ – Mohammad Zuhair Khan Dec 31 '18 at 6:12