For $x \in \mathbb{R}$ consider the series $F(x)= \sum\limits_{n \in \mathbb{Z} } \exp \left (- \left (x-\frac{n}{2} \right)^2 \right )$. I need to prove that there exists a constant $C$ such that $F(x) \leq C$ for all $x\in \mathbb{R}$.
I have already checked that $F(x) = F(x- \left \lfloor{x}\right \rfloor)$ for all $x \in \mathbb{R}$, so I can reduce the problem to finding an upper bound for all $x \in [0,1]$. And I assume that I need the monotony of $\exp$, but I am not sure how to proceed from here.
Let $x \in \mathbb{R}$. One has $$F(x)= \sum_{n \in \mathbb{Z}} \exp\left( -\dfrac{1}{4}(2x-n)^2 \right)$$
Substituting $k=\lfloor 2x \rfloor -n$, you get $$F(x)= \sum_{k \in \mathbb{Z}} \exp\left( -\dfrac{1}{4}(k+\lbrace 2x \rbrace)^2 \right), \quad \text{where } \lbrace 2x \rbrace = 2x - \lfloor 2x \rfloor \in [0,1[$$
So you can bound directly $F(x)$ by $$0 \leq F(x) \leq \sum_{k \in \mathbb{Z}} \exp\left( -\dfrac{1}{4}k^2 \right) = 1 + 2 \sum_{k =1}^{+\infty} \exp\left( -\dfrac{k^2}{4} \right)$$