Show that if $f \in L^1 \cap C$, then $\sum_{n \in \mathbb{Z}} f(x+n)$ exists and is uniformly continuous

I'm trying to prove the following statement:

Let $f \in L^1 \cap C$ such that there exists $A,B>0$ such that if $|x|>R$ we have that $$|f(x)| < \frac{B}{x^2}$$ Then, the function $$\sum_{n \in \mathbb{Z}} f(x+n)$$ exists

However, I'm not sure where to begin. I think I need to prove that the serie converges uniformly on $\mathbb{R}$ but with that translation over all $\mathbb{Z}$ I'm having trouble bounding this serie.

First I know what $f(x+n)$ is uniformly continuous on any bounded set hence it is continuous on $[-R,R]$ but does the bound in the proposition holds for all $n \in \mathbb{Z}$?

I thought about using the Weierstras criteria for the uniform convergence, i.efor each $n \in \mathbb{Z}$, we have that $|f(x+n)| < \frac{M}{(x+n)^2}$ for $x \notin [-R,R]$ but then I dont have a uniform bound on my serie.

Any help would be really appreciated!

You're just asked to show the function is defined first. So I'd start with that.

For any $x \in \mathbb{R}$, there exists sufficient large $N$ (depending on $x$) such that

\begin{align} \sum_{n \in \mathbb{Z}} |f(x+n)| &= \sum_{x \leq |N|} |f(x+n)| + \sum_{n > |N|}|f(x+n)| \\ & \leq \sum_{x \leq |N|} |f(x+n)| + \sum_{n> |N|} \frac{B}{(x+n)^2} \end{align}

This converges. Therefore $\sum f(x+n)$ converges.

To show continuity, you can take $x$ and $y$ close to each other and show that $\sum f(x+n) - \sum f(y+n) $ are close by using the same inequality above.

Uniform continuity follows from the fact that $F(x) = \sum f(x+n) $ is periodic. Since it's continuous on $[0,1]$ it's uniformly continuous on $[0,1]$. Since $F$ is periodic, it's uniformly continuous on all of $\mathbb{R}$.