Let $f$ be a Schwartz function on $\mathbb R$. I want to prove that
$$\sum_{n \in \mathbb Z}f(n) = \sum_{n \in \mathbb Z}\widehat{f}(n)$$
The beginning of my proof goes like this: the decay property of Schwartz functions ensures that both sums are absolutely convergent (for example, use the power test for series). Let $F(x) := \sum_{n \in \mathbb Z}f(x + n)$. Then $F(x)$ is a 1-periodic $C^\infty$-function, and so by the Dirichlet-Jordan Test, it converges absolutely uniformly everywhere to its Fourier expansion...and we proceed from here.
My question is this: how can we prove that $F(x)$ is $C^\infty$ (or even differentiable on $\mathbb R$)? It suffices to prove that $\sum_{n \in \mathbb Z} f'(x + n)$ converges uniformly on $\mathbb R$, but I am not sure how to do this.