Convergence of modified series with $e^{-x^2}$ Let $f(x) := e^{-x^2}$ and I want to check, if $\sum_{n=0}^{\infty} f(x-n)$ convergence to some function $F(x)$. Also I want to check type of convergence and if $F \in C^1$ and if it is Lipschitz.
So, it cannot be uniformly convergence, as sequence $f_n = f(x-n)$ isn't uniformly convergence to zero.
I think it's pointwise convergence, cause for some given fixed $x$ we can reject $k = 2 \lceil x \rceil$ firs elements, and we will have (in new sequence) $g_i(x) = f_{k+i}(x) < f(-i)$. It was finite number of elements, so nothing changed. $g_i$ converges by comparison test.
However, it's not correct solution. How can I prove pointwise convergence and find function $F'$?

If I'll find $F'(x)$, then it will be easy to check, if it's Lipschitz. (Lagrange theorem.) Just checking $|F'(x)| < M$, where $M$ is constant.
As it is pointwise convergence, I was thinking about $\sum f_n'(x)$, but it has to be uniformly convergence. It isn't.
So still I dont know, how to find $F'$ or show that $F'$ exists and function satisfy Lipschitz continuity.

$$\sum_{n=0}^\infty f(x-n)=\sum_{n=0}^\infty e^{-(x-n)^2}=e^{-x^2}\sum_{n=0}^\infty e^{2xn}e^{-n^2}$$
Apply now the n-th root test to the general term in the series
$$\sqrt[n]{e^{2xn}e^{-n^2}}=e^{2x}e^{-n}\xrightarrow[n\to\infty]{}0$$
so the series converges for any $\;x\in\Bbb R\;$ .
