calculus - Show that $sum_{m in mathbb{Z}} e^{-(x-m)^4}$ converges for all $x in mathbb{R}$ converges

how can I rigorously show that

$$\sum_{m \in \mathbb{Z}} e^{-(x-m)^4}$$ converges for all real $x$?

I am aware of convergence criteria for ordinary series, but not for $\sum_{m \in \mathbb{Z}}$. Does anybody have an idea?

In particular, I am also looking for $A,B>0$ such that

$$A \le \sum_{m \in \mathbb{Z}} e^{-(x-m)^4} \le B$$ for all $x \in \mathbb{R}.$

If you have any questions, please let me know.

Answer

For $x\in [0,1],$ note $|x-m|\ge |m|-|x|\ge |m|-1.$ Also $|m|-1 \ge |m|/2$ for $|m|\ge 2.$ So for any such $x$ your sum is less than $3 + 2\sum_{m=2}^{\infty}e^{-(|m|/2)^4},$ and that last series is convergent (by a mile).

The proof for other $x$ is similar, although it might be more relaxing to note that the sum as a function of $x$ is periodic.