I'm not really sure what topics exactly this falls under, so I apologize if I've misclassified this question.
There is a neat way of computing $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ using Fourier analysis: Compute the Fourier series of t^2$t^2$ extended 2π$2\pi$-periodically, which turns out to be
\frac{\pi^{2}}{3}+4\sum\sb {n=1}^{\infty}\frac{1}{n^{2}} http://latex.mathoverflow.net/png?%5Cfrac%7B%5Cpi%5E%7B2%7D%7D%7B3%7D%2B4%5Csum%5F%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%5E%7B2%7D%7D$$\frac{\pi^{2}}{3}+4\sum_{n=1}^{\infty}\frac{1}{n^{2}}$$
By Fejer's theorem (I think), the Fourier series around π$\pi$ converges, so we get an equation that can be solved for the zeta(2)$\zeta(2)$.
I think a similar approach can be taken for zeta(2k)$\zeta(2k)$ by taking t^{2k}$t^{2k}$ extended 2π$2\pi$-periodically, but all my attempts to do something like this for odd integers fail.
On the other hand, since {1/n^k}$1/n^k$ for k$k$ odd is in l^2$\ell^2$, there should be an L^2$L^2$ function that has the sequence as its Fourier coefficients. Can one be explicitly constructed? What if we allow the entries in the sequence to alternate, or let finitely many of them deviate from 1/n^k$1/n^k$?
Basically, I want to find an L^2$L^2$, 2π$2\pi$-periodic function whose Fourier coefficients would give a relatively straightforward computation of zeta(k)$\zeta(k)$ when k$k$ is odd.
