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Generalized geometric series

  1. Mar 23, 2012 #1
    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

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

    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. jcsd
  3. Mar 24, 2012 #2
    Correcting a very elementary integration error, I came up instead:

    [itex]\sqrt\pi=\lim_{R\to\infty}2\sqrt{R}\sum_{n=0}^{∞} \frac{(-1)^n R^n}{n!(2n+1)}[/itex]
     
  4. Mar 24, 2012 #3

    mathman

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    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.
     
  5. Mar 24, 2012 #4
    Yeah, the reason I'm disinterested in that one is that using it gives:

    [itex]\sqrt\pi = \lim_{R\to\infty} 2\sqrt R \frac{\sqrt\pi erf(R)}{2\sqrt R} = \sqrt\pi erf(\infty)=\sqrt\pi[/itex].

    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!
     
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