Does this weighted sum of Gaussians converge to $1/x^2$? By coincidence I discovered that a weighted sum of several Gaussians seems to converge to $1/x^2$. Now I wonder whether that's a well-known property, just a coincidence or if this could be proven.
This is a normal Gaussian with an additional weight factor $1/\sigma$:
$$
f(x, \sigma) = \frac{1}{\sigma} \cdot \frac{1}{\sigma \sqrt{2\pi} } e^{-\frac{1}{2}\left(\frac{x}{\sigma}\right)^2}
$$
If I now accumulate several of these weighted Gaussians with exponentially increasing $\sigma$, I get a curve which closely resembles $1/x^2$. The additional factor $\sqrt{3}$ was determined by trial and error and may not be the correct value.
$$
g(x, n) = \sqrt{3} \cdot \sum_{k=0}^n f\left(x, \frac{2^k}{n}\right)
$$
$$
\lim\limits_{n \to \infty} g(x, n) \stackrel{?}{=} \frac{1}{x^2}
$$
Does anybody have an idea how this could be proven?
 A: Actually, $\lim_{n\to\infty}g(x,n)$ doesn't exist, but $\lim_{n\to\infty}g(x,2^n m)$ exists for any integer $m>0$, and it depends on $m$; however, as $m$ changes, it oscillates around $1/x^2$ with very small amplitude. And, to make its "average value" exactly $1/x^2$, your $\sqrt{3}$ has to be replaced with $\sqrt{2\pi}\log 2$.
For $x\neq 0$ we have $$x^2 g(x,2^n m)=\sqrt\frac{3}{2\pi}\sum_{k=0}^{2^n m}4^{n-k}(mx)^2 e^{-4^{n-k}(mx)^2/2}\underset{n\to\infty}{\longrightarrow}\sqrt\frac6\pi G\left(4,\frac{(mx)^2}{2}\right),$$ where $G(a,b)=\sum\limits_{k=-\infty}^\infty(a^k b)e^{-a^k b}$ for $a>1$ and $b>0$.
Following this answer of mine literally, we obtain (here, $i$ is the imaginary unit) $$G(a,b)=\frac{1}{\log a}\sum_{n=-\infty}^{\infty}\Gamma(1+s_n)b^{-s_n},\qquad s_n=\frac{2n\pi i}{\log a}.\tag{*}\label{gammaseries}$$
As $b$ (i.e. $m$) changes, this oscillates around $1/\log a$, and the amplitudes for our $a=4$ are $$|\Gamma(1+s_{\pm 1})|\approx4.3\cdot10^{-3},\quad|\Gamma(1+s_{\pm 2})|\approx4.9\cdot10^{-6},\quad|\Gamma(1+s_{\pm 3})|\approx4.9\cdot10^{-9},\quad\ldots$$

Another way to get the formula \eqref{gammaseries}. Consider $G(a,a^x)$ as a function of $x\in\mathbb{R}$. It is smooth and has a period $1$, hence a convergent Fourier expansion $G(a,a^x)=\sum_{n\in\mathbb{Z}}c_n e^{-2n\pi ix}$, where
\begin{align*}
c_n&=\int_0^1 G(a,a^x)e^{2n\pi ix}\,dx\\
&=\sum_{k\in\mathbb{Z}}\int_0^1 a^{k+x}e^{-a^{k+x}+2n\pi ix}\,dx\\
&=\sum_{k\in\mathbb{Z}}\int_k^{k+1}a^t e^{-a^t+2n\pi it}\,dt\\
&=\int_{-\infty}^\infty a^t e^{-a^t+2n\pi it}\,dt\qquad\color{gray}{[a^t=z]}\\
&=\frac{1}{\log a}\int_0^\infty z^{s_n}e^{-z}\,dz=\frac{\Gamma(1+s_n)}{\log a}.
\end{align*}
