Fourier Series and the Riemann-Zeta Function

[MaGG]

Homework Statement

Use the Fourier series technique to show that the following series sum to the quantities shown:
$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$$
$$1+\frac{1}{3^2}+\frac{1}{5^2}+...=\frac{\pi^2}{8}$$

Homework Equations

I know the Riemann-Zeta function is
$$\zeta (m)=\sum_{n=1}^\infty \frac{1}{n^m}$$

The Attempt at a Solution

For the first series, I know that it's the evaluation of $\zeta (4)$. The problem is getting the derivation down. I've found an example of it being solved and proven at $\zeta (2)$ here: http://planetmath.org/?op=getobj&from=objects&name=ValueOfTheRiemannZetaFunctionAtS2 . The problem is, I don't really understand what to define as the initial function $f(x)$.

For the second series, I think I've found it to be:
$$\frac{1}{(2n-1)^2}$$
After that, I don't know how to go about solving it using the Fourier method. Any help would be greatly appreciated, thanks!

 

[e(ho0n3]

By "Fourier series technique", do you mean expand the 1/n⁴ or the 1/(2n - 1)² into its Fourier series?

 

[yasiru89]

A general method appears in the paper 'Recursive Formulas for [itex]\zeta(2k)[/tex] and [itex]L(2k-1)[/tex]' by Xuming Chen.<br /> <br /> Generally, try [itex]f(x) = x^{2k}[/tex] for your function (an even one on x here) in a Fourier expansion to get [itex]\zeta(2k)[/tex], and consider the derivative of the expansion to get [itex]L(2k-1)[/tex] (the L function here is the analogue of the Riemann zeta on odd numbers only).[/itex][/itex][/itex][/itex][/itex]

 

[solamon]

but still you ll not express

$\zeta(2k+1)$