It seems that equi-distant Gaussian bell curves almost add up to the constant function $1$ already for relatively small sigmas, i.e. $$ \lim_{s\rightarrow\infty} \sum_{n = -\infty}^{\infty} \frac{1}{\sqrt{2\pi}s} e^{-\frac{1}{2}(\frac{x-n}{s})^2} = 1 $$ converges for all $x$ very fast.
Already for $s=0.65$ the maximal deviation is $<0.001$.
Is this a special property of the Gauss function? Is there a simple argument for this strong approximation behaviour?
The Poisson summation formula says $$ \sum_{n = -\infty}^\infty f(n) = \sum_{k=-\infty}^\infty F(k) $$ where $$ F(y) = \int_{-\infty}^\infty f(x) e^{-2\pi i y x}\; dx $$ is the Fourier transform of $f$. In this case with $$f(x) = \frac{1}{\sqrt{2\pi} s} \exp(-(x-x_0)^2/(2 s^2))$$ we have $$ F(y) = \exp(-2\pi i y x_0) \exp(-2 \pi^2 s^2 y^2) $$ so $$ \frac{1}{\sqrt{2\pi} s} \sum_{n=-\infty}^\infty \exp(-(n-x_0)^2/(2 s^2)) = \sum_{k=-\infty}^\infty \exp(-2 \pi i k x_0) \exp(-2 \pi^2 s^2 k^2) $$ The $k=0$ term on the right is $1$, and if $s$ is not too small the other terms will be very small (e.g. if $s = 1$ the $k=\pm 1$ terms have absolute value $\exp(-2\pi^2) < 2.7 \times 10^{-9}$) and the series converges rapidly.