# Gamma expansion

• Mar 26th 2011, 09:32 AM
paolopiace
Gamma expansion
I know that:

$\displaystyle c(n) = \sqrt{\frac{2}{n}}\ \frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{n-1}{2})} = 1 - \frac{3}{4n} + O(n^{-2})$

and also that:

$\displaystyle \sqrt{\frac{n}{n-1}}\ c(n) = \sqrt{\frac{n}{n-1}}\ (1 - \frac{3}{4n} + O(n^{-2})) = (1 - \frac{1}{4n} + O(n^{-2}))$

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

Seeking directions/help. Thanks a bunch!
• Mar 26th 2011, 12:15 PM
Moo
Hello,

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

write the expansion for $\displaystyle \displaystyle \frac{1}{1-\frac 1n}$ (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 ($\displaystyle (1+x)^a=...$)
• Mar 26th 2011, 01:09 PM
paolopiace
Thanks, good idea but it seems it does not work (unless major mistakes):

Let u = 1/n. Then:

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

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

$\displaystyle \sqrt{1+x} = 1 + \frac{1}{2}x + O(x^2) = 1 + \frac{1}{2n} + O(n^{-2})$

Which means, the second term of the resulting series is the incorrect +1/(2n) instead of the wanted -1/(4n).
• Mar 26th 2011, 01:22 PM
Moo
And what if you compute $\displaystyle \displaystyle \left(1+\frac{1}{2n}+O(n^{-2})\right)\left(1-\frac{3}{4n}+O(n^{-2})\right)$ ? :D
• Mar 26th 2011, 01:26 PM
paolopiace
(Doh)

...thanks...