# Generalized geometric series

1. Mar 23, 2012

### alexfloo

I just sent some time dicking around with the MacLaurin expansion of exp(-z2) to derive a series expression for √π, by integrating term-by-term along the real line. I'm not really concerned with wether this is a useful or well-studied expression, I just thought it would be a fun exercise.

Since the individual term-wise integrals diverge, we have to stick with a limit. It came down to

$\frac{\sqrt{\pi}}{4}=\lim_{R\to\infty}\sqrt{R}\sum_{n=0}^{\infty}\frac{(-1)^nR^n}{(n-1)!}$

As R increases, the rate of convergence of the series decreases, so this is pretty useless unless we have a closed form for that series.

There's an obvious form in terms of the error function, but does anyone know of one that would actually lead to a nontrivial result?

2. Mar 24, 2012

### alexfloo

Correcting a very elementary integration error, I came up instead:

$\sqrt\pi=\lim_{R\to\infty}2\sqrt{R}\sum_{n=0}^{∞} \frac{(-1)^n R^n}{n!(2n+1)}$

3. Mar 24, 2012

### mathman

As you noted, you can get a closed form solution in terms of the error function, which gives the answer as R becomes infinite. I am not sure what you are looking for.

4. Mar 24, 2012

### alexfloo

Yeah, the reason I'm disinterested in that one is that using it gives:

$\sqrt\pi = \lim_{R\to\infty} 2\sqrt R \frac{\sqrt\pi erf(R)}{2\sqrt R} = \sqrt\pi erf(\infty)=\sqrt\pi$.

Basically it appears that my original expression was circular (i.e. trivial) in nature to begin with, but I was hoping to find a more useful expression. In any case, it seems like a dead end. Thanks anyways, though!