Suppose that $f(x)$ is a Schwarz function, which is defined here to be a function
that satisfies for every $c \in \mathbb{R}, n \in \mathbb{N}, |f^{n}(x)| = o(|x|^c)$,
then in a proof of the Poisson Summation Formula here, the author states that the function
$F(x)$ is $1$-periodic, where $F(x) = \sum_{n \in \mathbb{Z}}f(x+n)$, with the reason given being absolute convergence.
I'm imagining if $F(x)$ is convergent, then $F(x+n)=F(x)$ for $n \in \mathbb{Z}$ since we are simply displacing the series values by $n$, but ultimately the series is the same.
I do not understand however why $F(x)$ converges at all. Why does the absolute convergence of the series $\sum_{\mathbb{N}} |f(x+n)|$ imply the convergence of $F(x)$?
Proof of $F(x)$ converges:
since $|f(x)|=o(|x|^{-2})$ then $\sum_{n \in \mathbb{Z}}|f(x+n)|\leq\sum_{n \in \mathbb{Z}}|x+n|^{-2}$ which is finite thanks to the Basel problem unless $x+n=0$, in which case just remove the finite term from the left hand side of the summation. Since we know that an absolutely convergent series of complex numbers is convergent, then $F(x)$ is convergent.