It's a well known result that the Gaussian integral
$$\int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$$
evaluates to $\frac{\sqrt{\pi}}{2}$. This result can be obtained using double integrals with polar coordinates, among other things, but I'm particularly interested in evaluating this integral using power series.
The power series of $e^{-x^2}$ is
$$\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}$$
and integrating this, we get
$$\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(n!) (2n+1)}$$
This expression is our integral (and it also has a radius of convergence of infinity, so we know it’ll approximate our integral function everywhere), so to evaluate it, we plug in our bounds. The expression is 0 for the bound of 0, so we're just looking for the limit of this power series as it tends to infinity.
Using graphing software, with enough terms from the power series, I can see that the function does indeed spend a while at $\frac{\sqrt{\pi}}{2}$ before diverging.
So, how can I analytically prove that the limit of the power series
$$\lim_{x\to\infty} \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(n!) (2n+1)}$$
as $x$ approaches infinity is $\frac{\sqrt{\pi}}{2}$?Thanks for your help!