Proving $ \sum_{n=-\infty}^{\infty}(-1)^n e^{-\left(x - \frac{n}{2}\right)^2} = K\cos(2\pi x)$ I conjecture that:
$$ \sum_{n=-\infty}^{\infty}(-1)^n e^{-\left(x - \frac{n}{2}\right)^2} = K\cos(2\pi q x)$$
for some $K \in \mathbb{R}, q \in \mathbb{Z}$. Let $G(x)$ denote the left side. 
It's easy to show that $G(x + 1) = G(x)$ (by a shifting sum argument). 
Furthermore it can be shown that $G(-x) = G(x)$ by noting that for each summand 
$$(-1)^n e^{-\left(x- \frac{n}{2}\right)^2} $$
If we invert it 
$$(-1)^n e^{-\left(-x- \frac{n}{2}\right)^2}$$
Thats the same as 
$$(-1)^n e^{-\left(x+ \frac{n}{2}\right)^2}$$
And since the sum is over all natural numbers that implies this mapping is a bijection of terms in the sum to terms in the sum. 
But is the set of functions $\mathbb{C} \rightarrow \mathbb{C}$ that satisfy $ F(x+1) = F(x) , F(x) = F(-x)$  really equal to 
$K\cos(2\pi q x)$?
I worry that some pathological examples might arise that satisfy both equations but are excluded from this list. What other conditions should I chase? Trying to show that the second derivative is a multiple of the original function has not been very fruitful. I end up with the expression 
$$ G'' =  2\sum_{n=-\infty}^{\infty}(-1)^{n+1} \left(1 - 2\left( x - \frac{n}{2} \right)^2 \right) e^{-\left(x - \frac{n}{2}\right)^2} $$ 
(if we let $u = e^{x^2} G$ then I think this yields)
$$ (4x^2 - 2)e^{-x^2} U -4xe^{-x^2}U' + e^{-x^2} U'' = (e^{x^2}U)'' $$ 
But involving matrix exponentials to solve this seems like a mess. 
 A: We have from the Poisson summation formula that
$$f_p(x)=\sum_{n \in \mathbb{Z}} f(x+np) = \frac{1}{p}\sum_{n\in\mathbb{Z}}\hat{f}\left(\frac{n}{p}\right)e^{2i\pi\frac{n}{p}x}.$$
In our case $f(x)=e^{-x^2}-e^{-\left(x+\frac{1}{2}\right)^2}$ and $p=1$, which gives $\hat{f}(\xi)=\sqrt{\pi}e^{-\pi^2  \xi^2}(1-e^{\pi i \xi})$. Thus,
$$G(x)=f_{1}(x)=\sqrt{\pi}\sum_{n\in\mathbb{Z}}e^{-\pi^2 (2n+1)^2 }e^{(4n+2)i \pi x} = 2\sqrt{\pi}\sum_{n\geq 0} e^{-\pi^2 (2n+1)^2}cos((4n+2)\pi x).$$
Thus, $G(x)$ is not a multiple of cosine.
A: The conjecture is that the equation holds for $q=1$, which was supported in comments by numerical evidence. 
If this is true, then the continuous Fourier spectrum should exhibit a single peak at frequency $f =1$, and be zero elsewhere.
We already know from OP that the LHS $G(x)$ is periodic with period $1$ in $x$.
Consider a rational $f = p/r$ with natural numbers $p$ and $r$. 
The  Fourier spectrum G(f) of the LHS can then be computed, for any  $k \in \mathbb{Z}$, from (apart from a fixed normalization factor)
$$
G(f) = \int_{k r}^{(k+1)r} G(x) \cos(2\pi f x) dx
$$
since $r$ is the GCD of $1$ and $f$. 
Then it may also be computed as the average of several $(2 M)$ of these intervals:
$$
G(f) = \frac{1}{2M} \sum_{k=-M}^{M-1} \int_{k r}^{(k+1)r} G(x) \cos(2\pi f x) dx = \frac{1}{2M} \int_{-Mr}^{Mr} G(x) \cos(2\pi f x) dx
$$
Consider $f \ne 1$ (more generally, exclude all odd $f$ , (1,3,5,...)).
Compute $G(f)$, with some help by Wolfram alpha, as 
$$
G(f) = \lim_{M \to \infty}\frac{1}{2M} \int_{-Mr}^{Mr} G(x) \cos(2\pi f x) dx \\
= \lim_{M \to \infty}\frac{1}{2M} \sum_{n=-\infty}^{\infty}(-1)^n \int_{-\infty}^{\infty} e^{-\left(x - \frac{n}{2}\right)^2} \cos(2\pi f x) d x\\
= \lim_{M \to \infty}\frac{1}{2M} \sum_{n=-\infty}^{\infty}(-1)^n \sqrt\pi e^{-\pi^2 f^2} \cos(\pi n f)\\
= \lim_{M \to \infty}\frac{1}{2M} \sqrt\pi e^{-\pi^2 f^2}  (-1 + 2\lim_{m \to \infty} \sum_{n=0}^{m}(-1)^n \cos(\pi n f)  )\\
= \lim_{M \to \infty}\frac{1}{2M} \sqrt\pi e^{-\pi^2 f^2} \lim_{m \to \infty}   (-\cos(\pi (f + 1) (m + 1)) - \tan(\pi f/2) \sin(\pi (f +1) (m + 1)) )
$$ 
Now note that the inner limit $\lim_{m \to \infty} $ is bounded in absolute values by a finite $1 + |\tan(\pi f/2)|$. Hence 
$$
|G(f)| \leq   \lim_{M \to \infty}\frac{1}{2M} \sqrt\pi e^{-\pi^2 f^2} (1 + |\tan(\pi f/2)|) \to 0 
$$ 
Hence all Fourier components other that for $f=1$ disappear, which proves that the LHS $G(x)$ has the only frequency $f=1$, with $K=G(0)$, which in turn shows that it equals the RHS. $\qquad \Box$
