Through inspection, the following appears to be true:
$$\sum_{-\infty}^{\infty} (-1)^n e^{-(x-n)^2} = k\cos(\pi x)$$
where $k\approx3$. Even the case for $n=2$ appears to be a very good approximation for $x\in[-0.5, 0.5]$. However, I can't think of any reason why this would be true. Is this equality actually true and, if so, why?
Let $f_x(t) = \exp\left[i\pi(t-x)-(t-x)^2\right]$. Note that $f_0(t) = \exp(i\pi t - t^2)$ has Fourier transform $$\widehat{f}_0(\omega) = \int_{-\infty}^{\infty}f_0(t)e^{-i2\pi t\omega}\,dt = \sqrt{\pi}\exp\left[-\pi^2\left(\omega-\tfrac{1}{2}\right)^2\right].$$
Hence, the shifted version $f_x(t) = f_0(t-x)$ has Fourier transform $$\widehat{f}_x(\omega) = e^{-i2\pi x\omega}\widehat{f}_0(\omega) = \sqrt{\pi}\exp\left[-i2\pi x\omega -\pi^2\left(\omega-\tfrac{1}{2}\right)^2\right].$$
Then, the Poisson summation formula gives us $$\begin{align*}\sum_{n = -\infty}^{\infty}f_x(n) &= \sum_{k = -\infty}^{\infty}\widehat{f}_x(k) \\ \sum_{n = -\infty}^{\infty}\exp\left[i\pi(n-x)-(n-x)^2\right] &= \sum_{k = -\infty}^{\infty}\sqrt{\pi}\exp\left[-i2\pi xk -\pi^2\left(k-\tfrac{1}{2}\right)^2\right] \\ \sum_{n = -\infty}^{\infty}\exp\left[i\pi n -(n-x)^2\right] &= \sum_{k = -\infty}^{\infty}\sqrt{\pi}\exp\left[i\pi x -i2\pi xk -\pi^2\left(k-\tfrac{1}{2}\right)^2\right] \\ \sum_{n = -\infty}^{\infty}(-1)^ne^{-(x-n)^2} &= \sum_{k = -\infty}^{\infty}\sqrt{\pi}\exp\left[-i2\pi x(k-\tfrac{1}{2}) -\pi^2\left(k-\tfrac{1}{2}\right)^2\right] \end{align*}$$ where we multiplied both sides by $\exp(i\pi x)$ and simplified things.
Now, if we split the right side into the ranges $k = -\infty, \ldots, 0$ and $k = 1,\ldots,\infty$, make the substitution $k' = 1-k$ into the first sum and then combine things, we get
$$\begin{align*}&\sum_{k = -\infty}^{\infty}\sqrt{\pi}\exp\left[-i2\pi x(k-\tfrac{1}{2}) -\pi^2\left(k-\tfrac{1}{2}\right)^2\right] \\ = &\sum_{k = -\infty}^{0}\sqrt{\pi}\exp\left[-i2\pi x(k-\tfrac{1}{2}) -\pi^2\left(k-\tfrac{1}{2}\right)^2\right] + \sum_{k = 1}^{\infty}\sqrt{\pi}\exp\left[-i2\pi x(k-\tfrac{1}{2}) -\pi^2\left(k-\tfrac{1}{2}\right)^2\right] \\ = &\sum_{k' = 1}^{\infty}\sqrt{\pi}\exp\left[i2\pi x(k'-\tfrac{1}{2}) -\pi^2\left(k'-\tfrac{1}{2}\right)^2\right] + \sum_{k = 1}^{\infty}\sqrt{\pi}\exp\left[-i2\pi x(k-\tfrac{1}{2}) -\pi^2\left(k-\tfrac{1}{2}\right)^2\right] \\ = &\sum_{k = 1}^{\infty}\sqrt{\pi}\left[\exp\left(i2\pi x(k-\tfrac{1}{2})\right) + \exp\left(-i2\pi x(k-\tfrac{1}{2})\right)\right]\exp\left(-\pi^2(k-\tfrac{1}{2})^2 \right) \\ =& \sum_{k = 1}^{\infty}2\sqrt{\pi}\cos(2\pi x(k-\tfrac{1}{2}))\exp\left(-\pi^2(k-\tfrac{1}{2})^2 \right) \end{align*}$$
Therefore, we have $$\sum_{n = -\infty}^{\infty}(-1)^ne^{-(x-n)^2} = \sum_{k = 1}^{\infty}2\sqrt{\pi}\exp\left(-\pi^2(k-\tfrac{1}{2})^2 \right)\cos(2\pi x(k-\tfrac{1}{2})).$$
Now, it's not hard to see that the $k = 1$ term is $2\sqrt{\pi}e^{-\pi^2/4}\cos(\pi x)$, and the $k = 2$ term is $2\sqrt{\pi}e^{-9\pi^2/4}\cos(3\pi x)$, etc. Notice that $2\sqrt{\pi}e^{-9\pi^2/4} \approx 8 \times 10^{-10}$, and the terms for $k \ge 3$ are even smaller. Hence, the sum is approximately $2\sqrt{\pi}e^{-\pi^2/4}\cos(\pi x)$.
