TITLE: What is the limit function of the series $\sum_{n=1}^{\infty} \left(e^{-n^2}-e^{-(n+x)^2}\right)$?
QUESTION [6 upvotes]: Consider the series
$$\sum_{n=1}^{\infty}\left(e^{-n^2}-e^{-(n+x)^2}\right).$$
According to some theorems in my research, I found that it is convergent on $\mathbb{R}$, but I cannot find the limit function of this exactly. Anyone can help me to obtain this? thanks a lot.

REPLY [3 votes]: In addition to Dr. Wolfgang Hintze's answer. While the first part of the series has a mentioned closed form in terms of theta function, we can find an integral form for the second part, but only for $$x>0$$
Represent the series as:
$$\sum_{n=1}^\infty e^{-(n+x)^2}=e^{-x^2} \sum_{n=1}^\infty e^{-n^2} e^{-2nx}$$
We use the following integral representation:
$$e^{-n^2}=\frac{1}{2 \sqrt{\pi}} \int_{-\infty}^\infty e^{-t^2/4} e^{int} dt$$
We obtain:
$$\sum_{n=1}^\infty e^{-(n+x)^2}= \frac{e^{-x^2}}{2 \sqrt{\pi}} \int_{-\infty}^\infty e^{-t^2/4} dt \sum_{n=1}^\infty e^{-(2x-it)n}$$
Using the geometric series formula, and getting rid of the imaginary part, we have finally:

$$\sum_{n=1}^\infty e^{-(n+x)^2}= \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4}  \frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 } dt$$

For $x<0$ the series inside the integral diverges, and for $x=0$ the integral after summation diverges, which is why we needed $x>0$.
However, $f(0)$ is trivial and for $f(-x)$ we can use Dr. Wolfgang Hintze's result.