# An upper bound for an exponential series

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.

• It's trivial from here on. You've already shown you only need to show an upper bound for $x \in [0,1]$. And note that the maximum of $e^{-(x-t)^2}$ on $[0,1]$ is $1$ if $t \in [0,1]$, $e^{-(1-t)^2}$ for $t > 1$ and $e^{-(t)^2}$ for $t < 0$, which is a sum which converges (you can check this by the direct comparison test).
– Anon
Dec 28, 2022 at 17:43
• As always(?), virtually all you need about the exponential function is $\exp x\ge 1+x$ Dec 28, 2022 at 21:16

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 $$0 \leq F(x) \leq \sum_{k \in \mathbb{Z}} \exp\left( -\dfrac{1}{4}k^2 \right) = 1 + 2 \sum_{k =1}^{+\infty} \exp\left( -\dfrac{k^2}{4} \right)$$
• Nice way to do it. May be, you could add that this is just $\vartheta _3\left(0,\frac{1}{\sqrt[4]{e}}\right)$ Dec 29, 2022 at 10:28