Consider the series∑n=1∞(e−n2−e−(n+x)2

Question

disgynna0vv · Accepted Answer

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} (\exp (- n^{2}) - \exp (- {(n + x)}^{2}))

Symmetry:
Define

g (x) = f (x - \frac{1}{2}) + \frac{1}{2}

then

g (x)

is antisymmetric:

g (- x) = - g (x)

Hence follows

f (- x) = - 1 - f (x - 1)

(1)
2) Assymptotic values

a = f (x \to + \infty) = \sum_{n = 1}^{\infty} \exp (- n^{2}) = \frac{1}{2} (v_{3} (0, \frac{1}{e}) - 1) \tilde{=} 0.386319

where

v_{3}

is a thetea function.
And, using (4)

f (x \to - \infty) = - 1 - a \tilde{=} - 1.386319

3) Integer values of x
lead to finte sums due to cancellations

f (0) = 0

f (1) = \exp (- 1^{2}) + \exp (- 2^{2}) + \exp (- 3^{2}) + \dots - \exp (- {(1 + 1)}^{2}) = \exp (- 1^{2})

f (2) = \exp (- 1^{2}) + \exp (- 2^{2})

and generally,

f (k) = \sum_{n = 1}^{k} \exp (- n^{2}), k = 1, 2, \dots

and by (1)

f (- k) = - 1 - f (k - 1) = - 1 - \sum_{n = 1}^{k - 1} \exp (- n^{2}), k = 1, 2, \dots

Consider the series \sum_{n=1}^\infty(e^{-n^2}-e^{-(n+x)^2}