What is the limit function of the series $\sum_{n=1}^{\infty} \left(e^{-n^2}-e^{-(n+x)^2}\right)$? 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.
 A: This is not a complete solution but a collection of interesting partial results.
Let the sum in question be
$$f(x) = \sum _{n=1}^{\infty } \left(\exp \left(-n^2\right)-\exp \left(-(n+x)^2\right)\right)\tag{1}$$ 
1) Symmetry:
Define 
$$g(x) = f\left(x-\frac{1}{2}\right)+\frac{1}{2}\tag{2}$$ 
then $g(x)$ is antisymmtric: 
$$g(-x) = - g(x)\tag{3}$$.
Hence follows
$$f(-x) = - 1 - f(x-1)\tag{4}$$
2) Asymptotic values
$$a = f(x\to +\infty) = \sum _{n=1}^{\infty } \exp \left(-n^2\right) = \frac{1}{2} \left(\vartheta _3\left(0,\frac{1}{e}\right)-1\right) \simeq 0.386319\tag{5}$$
where $\vartheta _3$ is a Jacobi theta function.
And, using (4),
$$f(x\to -\infty) = -1 - a \simeq -1.386319$$
3) Integer values of $x$
lead to finite sums due to cancellations
$$f(0) = 0$$
$$f(1) = \exp \left(-1^2\right)+\exp  \left(-2^2\right) +\exp  \left(-3^2\right)+ ...\\ - \exp \left(-(1+1)^2\right)-\exp  \left(-(1+2)^2\right) - ... = \exp \left(-1^2\right)$$
$$f(2) = \exp \left(-1^2\right)+\exp  \left(-2^2\right) +\exp  \left(-3^2\right) +...\\ - \exp \left(-(2+1)^2\right)-\exp  \left(-(2+2)^2\right) - ... = \exp \left(-1^2\right)+\exp  \left(-2^2\right)$$
$$...$$ 
and generally,    
$$f(k) = \sum_{n=1}^k \exp \left(-n^2\right), k = 1, 2, ...\tag{6a}$$
and by (4)
$$f(-k)= -1-f(k-1)= -1 -\sum_{n=1}^{k-1} \exp \left(-n^2\right), k = 1, 2, ...\tag{6b}$$
A: 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.
