For a function defined by the series
$$ f(x) = \sum_{n \in \mathbb{Z}} a_n e^{i n x }, $$
we can infer its smoothness by the decay rate of the coefficients $a_n$.
However, how about a more general series
$$ g(x) = \sum_{n \in \mathbb{Z}} a_n e^{i \omega_n x } . $$
Here $\omega_n $ are real but not integral. More specifically, let us assume that $\omega_n \sim n $.
There should be some connection between the smoothness of $g$ and the decay rate of $a_n$, right?
On
Let $$f(x)=\sum_n a_n e^{i\omega_n x}, \qquad \sum_n |a_n| < \infty, \qquad \lim \inf |\omega_n-\omega_{n-1}| > 0$$
If for every $m$, $\sum_n |a_n \omega_n^m| < \infty$ then $f \in C^\infty$ and all its derivatives are bounded.
Conversely if $f \in C^\infty$ and all its derivatives are bounded then for every $k$ $$f_k (x) = f(x) e^{-x^2/(2 k^2)} =\frac{1}{2\pi}\int_{-\infty}^\infty \hat{f_k}(\omega) e^{i \omega x}d\omega, \qquad\hat{f_k}(\omega) = \sum_n a_n k\sqrt{2\pi} e^{- (\omega-\omega_n)^2 k^2/2} $$ is smooth and fast decreasing as well as all its derivatives.
Therefore for every $m$, $\| f_k^{(m)}\|_{L^1} < \infty \implies \sup_\omega |\hat{f_k}(\omega) (i\omega)^m| < \infty$ and hence $\lim_{n \to \infty}a_n\omega_n^m=0$
You can apply DCT repeatedly to get such a result. Of course $\sum|a_n| <\infty$ implies continuity. Suppose $\sum |a_n \omega_n| <\infty$. Using the inequality $|1-e^{i\omega_n t}| \leq |\omega_n t|$ we see that $f$ is differentiable, and so on.