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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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This is function series but not a power-series - please fix the tag. – AD. May 17 '12 at 17:05
The cubic term pretty much guarantees that you're not going to find a result in terms of known special functions. As far as I am aware, there are no nontrivial theta function identities involving a $q^{n^{3}}$ term (which is $e^{-k^{3}}$ in this case. – deoxygerbe May 17 '12 at 17:36
Look up the Euler Maclaurin formula....Mathematica can do the integral, and it involves the erf function. – PeterR May 17 '12 at 18:42
@PeterR: Notice the integral $\int_1^\infty dz\, e^{-z(x-z)^2}$, is not a standard integral. – user26872 Jun 22 '12 at 14:36
up vote 3 down vote accepted

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

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How did you get \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^{5/2}}\right)\right) \\ &=& \exp\left( -\frac{z^2}{2} - O\left(\frac{1}{k}\right)\right)? \end{eqnarray*} – Mark Jul 14 '12 at 14:48
@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. – user26872 Jul 14 '12 at 18:05
@Mark: I found a minor error, $O(1/k^{5/2}) \rightarrow O(1/k^2)$. It has no effect on the result. – user26872 Jul 14 '12 at 18:21

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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How did you solve this integral? I can not do... – Mark May 19 '12 at 22:07
@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. – Sasha May 20 '12 at 5:04

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