
Gamma expansion
I know that:
$\displaystyle
c(n) = \sqrt{\frac{2}{n}}\ \frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{n1}{2})} =
1  \frac{3}{4n} + O(n^{2})
$
and also that:
$\displaystyle
\sqrt{\frac{n}{n1}}\ c(n) = \sqrt{\frac{n}{n1}}\ (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!

Hello,
$\displaystyle \displaystyle \sqrt{\frac{n}{n1}}=\sqrt{\frac{1}{1\frac 1n}}$
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 ($\displaystyle (1+x)^a=...$)

Thanks, good idea but it seems it does not work (unless major mistakes):
Let u = 1/n. Then:
$\displaystyle
\left(\frac{1}{1u}\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).

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
