I've been looking at Henri Darmon's lectures notes for modular forms, and in lecture 3 he proves the Poisson summation formula, but one step of the proof is that $$\int_0^1\sum_{m\in\mathbb{Z}}f(x+m)e^{-2\pi i nx}dx=\sum_{m\in\mathbb{Z}}\int_0^1f(x+m)e^{-2\pi i nx}dx.$$ In order to justify this, I was hoping to show that the series converges uniformly. It should be noted that we're assuming that $f$ is a Schwarz function, meaning that for all $N\geq1$, $|f(x)|\ll|x|^{-N}$ as $x\to\infty$.
My thought was to try a direct approach $$|\sum_{m\in\mathbb{Z}}f(x+m)e^{-2\pi i n x}-\sum_{-M}^Mf(x+m)e^{-2\pi i nx}|\\=|\sum_{-\infty}^{-M}f(x+m)e^{-2\pi i n x}+\sum_{M}^\infty f(x+m)e^{-2\pi i nx}|\\\leq\sum_{-\infty}^{-M}|f(x+m)|+\sum_{M}^\infty|f(x+m)|.$$ Now my hope at this point was to exploit the Schwarzness of $f$ to get that this is at most $$\sum_{-\infty}^{-M}|x+m|^{-N}+\sum_{M}^\infty|x+m|^{-N}$$ which should hold assuming that $M$ is large enough, and $|x|\ll M$. I thought maybe some clever choice of $N$ could help me get a term which goes to zero, and doesn't depend on $x$, but I don't see how to do it.
EDIT: So right after posting this I realized that as $M\to\infty$ both of the sums should go to zero, and this should give me the result I want, right?