1. ## Fourier series

Hey guys.
I wrote the problem in word, it's much easier for me

2. Originally Posted by asi123
The problem is the other one, $(x^3)$, how can I get $\sum_{n=1}^\infty\frac1{n^3}$ from that?
You'll be famous if you succeed. $\sum_{n=1}^\infty\frac1{n^3}$ is Apéry's constant, for which there is no known closed form. Even less is known about $\sum_{n=1}^\infty\frac1{n^5}$.

3. Originally Posted by Opalg
You'll be famous if you succeed. $\sum_{n=1}^\infty\frac1{n^3}$ is Apéry's constant, for which there is no known closed form. Even less is known about $\sum_{n=1}^\infty\frac1{n^5}$.
Well, the question is find the sum of the series $\sum_{n=1}^\infty\frac1{n^5}$, are you sure that can not be done?

4. Originally Posted by asi123
Well, the question is find the sum of the series $\sum_{n=1}^\infty\frac1{n^5}$, are you sure that can not be done?
He is sure. And so am I.

Some research for you to do:

Riemann zeta function.

$\zeta(2n)=\frac{(-1)^{k+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$.

Very little is known about $\zeta(2n + 1)$ apart from what has already been mentioned for the case n = 1.

5. Originally Posted by mr fantastic
He is sure. And so am I.

Some research for you to do:

Riemann zeta function.

$\zeta(2n)=\frac{(-1)^{k+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$.

Very little is known about $\zeta(2n + 1)$ apart from what has already been mentioned for the case n = 1.
Just in case the original poster want to see $\zeta(2n)$ derivation, it is here.

6. Originally Posted by asi123
Hey guys.
I wrote the problem in word, it's much easier for me
I think you might need to combine the Fourier series using the convolution theorem for Fourier series, though if it will give anything usefull I don't know.

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