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


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. Pongrácz Dec 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 $(1-1/16)*zeta(4)=15/16*zeta(4)=pi^4 /96$

  • $\begingroup$ $\LaTeX \text{ Tip}:$ Use \zeta, \pi and \frac {a}{b} to obtain $\zeta$, $\pi$ and $\frac ab$ $\endgroup$ – Mohammad Zuhair Khan Dec 31 '18 at 6:12

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