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 $t^2$ extended $2\pi$-periodically, which turns out to be
$$\frac{\pi^{2}}{3}+4\sum_{n=1}^{\infty}\frac{1}{n^{2}}$$
By Fejer's theorem (I think), the Fourier series around $\pi$ converges, so we get an equation that can be solved for the $\zeta(2)$.
I think a similar approach can be taken for $\zeta(2k)$ by taking $t^{2k}$ extended $2\pi$-periodically, but all my attempts to do something like this for odd integers fail.
On the other hand, since $1/n^k$ for $k$ odd is in $\ell^2$, there should be an $L^2$ 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 $1/n^k$?
Basically, I want to find an $L^2$, $2\pi$-periodic function whose Fourier coefficients would give a relatively straightforward computation of $\zeta(k)$ when $k$ is odd.
Best Answer
Well, I wish you the best. But I don't think you are going to succeed for two reasons:
(1) Proving almost anything about $\zeta(2k+1)$ is hard.
(2) There are nice formulas for $\sum_{n=1}^{\infty} \cos (n \theta)/n^{2k}$ and for $\sum_{n=1}^{\infty} \sin (n \theta)/n^{2k+1}$. There are not particularly nice formulas for $\sin$ with even powers or $\cos$ with odd. One way to think about this is that the imaginary part of $\log (1-e^{i \theta})$ has a simple formula but the real part does not.