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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)?

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    @PeterR: Notice the integral $\int_1^\infty dz\, e^{-z(x-z)^2}$, is [not a standard integral](http://www.wolframalpha.com/input/?i=integrate+e%5E%28-k%28x-k%29%5E2%29%2C+%7Bk%2C1%2Cinfty%7D).2012-06-22

2 Answers 2

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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.

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 $\begin{eqnarray*} \log \cos^2 \epsilon &=& \log\left[\left(1-\frac{\epsilon^2}{2}+\frac{\epsilon^4}{24} + O(\epsilon^6)\right)^2\right] \\ &=& \log\left(1-\epsilon^2 + \frac{\epsilon^4}{3} + O(\epsilon^6)\right) \\ &=& \left(-\epsilon^2+\frac{\epsilon^4}{3}\right) - \frac{1}{2}\left(-\epsilon^2+\frac{\epsilon^4}{3}\right)^2 + O(\epsilon^6) \\ &=& -\epsilon^2 - \frac{\epsilon^4}{6} + O(\epsilon^6). \end{eqnarray*}$

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    @Mark: Hi Mark, I did a series expansion of $\log g(k+z/\sqrt{2k})$ about $k=\infty$ (or $1/k = 0$) and convinced myself that the higher order terms are suppressed. I'll add something about this to the answer.2012-07-14
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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:

enter image description here

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.

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    @Mark Regroup arguments of exponentials $-k (x-k)^2 + i \omega x = -k \left( x - k - i \frac{\omega}{2k} \right)^2 + i k \omega - \frac{\omega^2}{4k}$, and then see answers to [this question](http://math.stackexchange.com/questions/123961/on-the-integral-int-infty-infty-e-x-ti2-dx/123966#123966).2012-05-20