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Math Help - Gamma expansion

  1. #1
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    Gamma expansion

    I know that:

    <br />
c(n) = \sqrt{\frac{2}{n}}\ \frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{n-1}{2})} =<br />
1 - \frac{3}{4n} + O(n^{-2})<br />

    and also that:

    <br />
\sqrt{\frac{n}{n-1}}\ c(n) = \sqrt{\frac{n}{n-1}}\ (1 - \frac{3}{4n} + O(n^{-2})) = (1 - \frac{1}{4n} + O(n^{-2}))<br />

    Problem is: I am not able to prove the very last equality. That's all what I need.

    Seeking directions/help. Thanks a bunch!
    Last edited by paolopiace; March 26th 2011 at 11:35 AM. Reason: Simplifying
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  2. #2
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    Hello,

    \displaystyle \sqrt{\frac{n}{n-1}}=\sqrt{\frac{1}{1-\frac 1n}}

    write the expansion for \displaystyle \frac{1}{1-\frac 1n} (you know, the beginning of the Taylor series... I don't remember what's the exact name)
    and then for the square root, consider it as power 1/2 and make another expansion ( (1+x)^a=...)
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    Thanks, good idea but it seems it does not work (unless major mistakes):

    Let u = 1/n. Then:

    <br />
\left(\frac{1}{1-u}\right)^{1/2} = \left(\sum_{j=0}^\infty u^j\right)^{1/2} = (1 + 1/n + O(n^{-2})\ )^{1/2}<br />

    Let  x = (1/n + O(n^{-2})) Then:

    <br />
\sqrt{1+x} = 1 + \frac{1}{2}x + O(x^2) = 1 + \frac{1}{2n} + O(n^{-2})<br />

    Which means, the second term of the resulting series is the incorrect +1/(2n) instead of the wanted -1/(4n).
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    And what if you compute \displaystyle \left(1+\frac{1}{2n}+O(n^{-2})\right)\left(1-\frac{3}{4n}+O(n^{-2})\right) ?
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    ...thanks...
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