When we do Fourier analysis, we don't always get convergent series. A classic example comes from considering the Sawtooth function. It has Fourier Coefficients $$s(x) = \frac{1}{2} + \sum_{n \neq 0} \frac{1}{2 \pi i n} e^{-2 \pi i n x}$$ Then $$s'(x) = 1 - \sum_{n \in \mathbb{Z}} e^{-2 \pi i n x}$$ Which of course doesn't converge. However, if we tack on a smooth cutoff function, then $s'(x) = 1$ at all the non-integers, exactly as we would expect. Indeed, this continues onward, since $$s''(x) = \sum_{n \in \mathbb{Z}} 2 \pi i n e^{-2 \pi i n x}$$ Which we expect to be zero, and is indeed zero with the added smooth cutoff function.
Let's take a slightly more interesting example. Take the Fourier coefficients $$2i \sum_{n=0}^\infty e^{-2 i (n+1) x} $$ Tacking on a smooth cutoff function allows us to recover that this function is $\cot(x)$.
A second method we can apply is complex integration. Take the contour from $-1/2 - i \infty$ up to $-1/2 + i \infty$ of the series, and we get back $\cot(x)$. More precisely, we have that $$\cot(x) = \frac{1}{\pi} \int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{e^{-2 i (n+1) x}}{e^{-2 \pi i n }-1} dn $$ This method also makes it clear that this relation plays well with point-wise derivatives, since $$\frac{d}{dx} \frac{1}{\pi} \int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{e^{-2 i (n+1) x}}{e^{-2 \pi i n }-1}dn = \frac{1}{\pi} \int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{d}{dx} \frac{e^{-2 i (n+1) x}}{e^{-2 \pi i n }-1} dn $$
Question
My question boils down to-- are there any general relationships between divergent series summation and Fourier series? IIRC there is a relationship between Cesaro's summation and Fourier series (something like every integral function has Fourier coefficients that are Cesaro summable?). Thus, I'm mainly interested in extensions of that case (how do the Fourier Coefficients the class of distributions interact with divergent series summation? What about cases where the Fourier coefficients have faster-than-polynomial growth?)
