# Evaluate $\sum^\infty_{n=1} \frac{1}{n^4}$using Parseval's theorem (Fourier series)

Evaluate $\sum^\infty_{n=1} \frac{1}{n^4}$using Parseval's theorem (Fourier series).

I have , somehow, to find the sum of $\sum_{n=1}^\infty \frac{1}{n^4}$ using Parseval's theorem.

I tried some things that didn't work so I won't post them.

Can you please explain me how do I find the sum of this series using Parseval's identity?

Thanks

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I suppose you know this, but you should find a function whose Fourier coefficients are ~ $\frac{1}{n^2}$ –  Poppy Jan 25 at 15:14
You have it solved here: math.cmu.edu/~bobpego/21132/fourierexample.pdf –  Poppy Jan 25 at 15:21

Let $f(x)=x^2$ for $x\in(-\pi,\pi)$. Computing the Fourier coefficients gives

$$a_n=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 e^{i n x} dx=\frac{2 \cos(\pi n)}{n^2}=2\frac{(-1)^n}{n^2}$$

for $n\in\mathbb{Z}$, $n\not=0$, and $a_0=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 dx=\frac{\pi^2}{3}$.

Therefore $|a_n|^2=\frac{4}{n^4}$ for $n\in\mathbb{Z}$, $n\not=0$ and $|a_0|^2=\frac{\pi^4}{9}$.

By Plancherel/Parseval's theorem,

$$\frac{\pi^4}{9}+8\sum_{n=1}^\infty \frac{1}{n^4}=\sum_{n=-\infty}^\infty |a_n|^2=\frac{1}{2\pi}\int_{-\pi}^\pi x^4 dx=\frac{\pi^4}{5}$$

Simplifying, this gives

$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{8}\left(\frac{1}{5}-\frac{1}{9}\right)=\frac{\pi^4}{90}$$

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Thank you very much .. one question: why is $n$ starts at $-\infty$ and not at $1$? –  Billie Jan 25 at 15:36
You need to consider all Fourier coefficients, that includes those with negative indices. –  Your Ad Here Jan 25 at 15:37
Thank you very much!!!!!!! –  Billie Jan 25 at 15:37
Hint: Look at $f(x) = x^2$ on an interval around $0$