What's the sum of $\sum_{k=1}^{\infty} e^{-k(x-k)^{2}}$? Absolute convergence and uniform convergence are easy to determine for this power series. What could be a possible approach to find the sum of this series 
$\sum_{k=1}^{\infty} e^{-k(x-k)^{2}}$
(the sum or an estimate of the sum)?
 A: 
This is really meant to be a comment, as the closed form for the series is unlikely. It studies the Fourier image of the function, defined by the series.

Clearly the series is absolutely convergent for all real $x$. Here is a plot of the series:

Oscillations suggest looking at the Fourier transform. Doing so, formally, term-wise gives:
$$
    \int_{-\infty}^\infty \mathrm{e}^{-k (x-k)^2} \mathrm{e}^{i \omega x} \mathrm{d} x = \sqrt{\frac{\pi}{k}} \mathrm{e}^{i \omega k}  \exp\left(-\frac{\omega^2}{k}\right)
$$
Thus, formally:
$$
  \mathcal{F}(\omega) = \sum_{k=1}^\infty \sqrt{\frac{\pi}{k}} \mathrm{e}^{i \omega k}  \exp\left(-\frac{\omega^2}{4 k}\right)
$$
As it is easy to notice, the series is divergent for $\omega = 2 \pi n$. It can be, however, turned into well defined function for all other real $\omega$:
$$\begin{eqnarray}
   \mathcal{F}(\omega) &=& \sum_{k=1}^\infty \sqrt{\frac{\pi}{k}} \mathrm{e}^{i \omega k}  \exp\left(-\frac{\omega^2}{4k}\right) \\
   &=& \sum_{k=1}^\infty \sqrt{\frac{\pi}{k}} \mathrm{e}^{i \omega k} + \sum_{k=1}^\infty \sqrt{\frac{\pi}{k}} \mathrm{e}^{i \omega k} \left(\exp\left(-\frac{\omega^2}{4k}\right) - 1\right) \\
   &=& \sqrt{\pi} \cdot \operatorname{Li}_{1/2}\left(\mathrm{e}^{i \omega}\right) + 2 \sum_{k=1}^\infty \sqrt{\frac{\pi}{k}} \mathrm{e}^{i \omega k} \exp\left(-\frac{\omega^2}{8 k}\right) \sinh\left(\frac{\omega^2}{8 k}\right)
\end{eqnarray}
$$
The unevaluated series is now absolutely convergent.
A: The approach here is heuristic.
We find an asymptotic formula for the sum for large $x$, Eq. (1) below.
Let $f(x) = \sum_{k=1}^\infty f_k(x)$ where $f_k(x) = e^{-k(x-k)^2}$.
Each term $f_k(x)$ is an unnormalized Gaussian distribution with mean $k$ and standard deviation $\sigma_k = 1/\sqrt{2 k}$.
For $k \ge 18$ we find $6\sigma_k \le 1$, that is, the distance between the means of two adjacent Gaussians is six or more standard deviations.
Thus, for large $x$ the function is a sum of narrow, well separated Gaussian spikes whose width decreases as $1/\sqrt{2 k} \approx 1/\sqrt{2 x}$.
Notice that $\cos^2\pi x$ has almost the right behavior, but the width is wrong.
A reasonable ansatz is $g(x) = (\cos^2 \pi x)^{h(x)}$, where $h(x)$ modulates the width of the spikes.
In fact, if we expand
$$g(x) = (\cos^2 \pi x)^{x/\pi^2}$$
about $x = k$ we find
$$g(x) \sim e^{-k(x-k)^2}.$$
Let's study this expansion in a little detail. 
Let $z=(x-k)/\sigma_k$.
Then,
$$\begin{eqnarray*}
g(z) &=& \exp\left(
-\frac{z^2}{2} - \frac{z^3}{2\sqrt{2}k^{3/2}} - \frac{\pi^2 z^4}{24k} + O\left(\frac{1}{k^{2}}\right)\right) \\
&=& \exp\left(
-\frac{z^2}{2} - O\left(\frac{1}{k}\right)\right),
\end{eqnarray*}$$
so in the limit we have a normal distribution with the appropriate mean and width.
Thus, for large $x$,
$$\begin{equation*}
\sum_{k=1}^\infty e^{-k(x-k)^2} \sim (\cos^2 \pi x)^{x/\pi^2}.\tag{1}
\end{equation*}$$
I tried to find such a solution from @Sasha's $\mathcal{F}(\omega)$ but had no luck.
It is likely something like this can be found by inverting $\mathcal{F}(\omega)$ in the appropriate limit.
Here's a plot of the sum and fit.

Figure 1. Plot of the sum (black) and fit (red).

Addendum: Series for $\log g(x)$
Let $x = k+z/\sqrt{2k}$ and expand about $k=\infty$, 
$$\begin{eqnarray*}
\log g(k+z/\sqrt{2k}) &=& \frac{k+z/\sqrt{2k}}{\pi^2}
\log \cos^2\pi\left(k+\frac{z}{\sqrt{2k}}\right) \\
&=& \frac{k+z/\sqrt{2k}}{\pi^2}
\log \cos^2 \frac{\pi z}{\sqrt{2k}}
    \hspace{10ex} (\textrm{sum formula for cosine, use }k\in\mathbb{Z}) \\
&=& \frac{k+z/\sqrt{2k}}{\pi^2}
\left[
-\left(\frac{\pi z}{\sqrt{2k}}\right)^2
- \frac{1}{6} \left(\frac{\pi z}{\sqrt{2k}}\right)^4
+ O\left(\frac{1}{k^3}\right)
\right] \\
&=&
-\frac{z^2}{2} - \frac{z^3}{2\sqrt{2}k^{3/2}} - \frac{\pi^2 z^4}{24k} + O\left(\frac{1}{k^{2}}\right).
\end{eqnarray*}$$
Notice that
$$\begin{eqnarray*}
\log \cos^2 \epsilon &=& \log\left[\left(1-\frac{\epsilon^2}{2}+\frac{\epsilon^4}{24} + O(\epsilon^6)\right)^2\right] \\
&=& \log\left(1-\epsilon^2 + \frac{\epsilon^4}{3} + O(\epsilon^6)\right) \\
&=& \left(-\epsilon^2+\frac{\epsilon^4}{3}\right)
- \frac{1}{2}\left(-\epsilon^2+\frac{\epsilon^4}{3}\right)^2 + O(\epsilon^6) \\
&=& -\epsilon^2 - \frac{\epsilon^4}{6} + O(\epsilon^6).
\end{eqnarray*}$$
