Let's start with a continuous function $f : \mathbb R \to \mathbb R$ with enough decay to make $$F(x) := \sum_{k \in \mathbb Z} f(x+k)$$ converge absolutely and uniformly (side question: does in this case absolute convergence imply uniform convergence?). Then $F$ is a $1$-periodic continuous function and we are allowed to change the summation with integration in the process of finding the Fourier coefficients of $F$ and it turns out $$a_n := \int_0^1 F(x)\exp(-2\pi i nx) dx = \hat f(n)$$ where $$\hat f(y) := \int_{\mathbb R} f(x) \exp(-2 \pi i xy) dx$$ is the Fourier transform of $f$.
Now in Fourier coefficients of smooth functions behave like Schwartz functions? I learnt that the smoothness of $F$ is equivalent to having a faster than any polynomial decay of the $a_n$. Does this give us the following equivalence $$ F \in C^\infty \quad \Leftrightarrow \quad f \in S(\mathbb R)$$ since the Fourier transform is an automorphism on $\mathcal S(\mathbb R)$? With $S(\mathbb R)$ I mean the Schwartz space of functions $g: \mathbb R \to \mathbb R$.