I have some troubles with the following series $$\sum^\infty _{n=0} x^ {n^2}$$
I'm suppose to show that this series is equivalent when $x$ approaches $1$ and $x <1$ to $$\frac{G}{\sqrt{1-x}}$$ where G = $$\int_0^{\infty}e^{-t^2}dt$$
I tried to use the series expansion of exponential function and then interchange the series and the integral but it doesn't work for obvious reasons .
Have you some tips ?
If $\ p:=1-x,\ $ then $\ x^{n^2} = (1-p)^{n^2} \approx e^{-n^2p}\ $ and $\ S:=\sum_{n=0}^\infty x^{n^2} \approx \sum_{n=0}^\infty e^{-n^2p} $ but also $\ \sum_{n=0}^\infty e^{-n^2p} \approx \frac1{\sqrt{p}}\int_0^\infty e^{-t^2} dt = \frac{\sqrt{\pi}/2}{\sqrt{1-x}}. $ A bit more work gives us that $\ S \sim \frac{\sqrt{\pi}/2}{\sqrt{1-x}}\ $ as $\ x\to 1^-.$
For example, if $\ x = 1 - 10^{-10},\ $ then the integral is $ \approx 88623.1925$ while $\ \frac{\sqrt{\pi}/2}{\sqrt{1-x}} \approx 88622.6925. $