Krdytkyu

Show $frac{Gamma((n-1)/2)}{Gamma(n/2)} approx frac{sqrt{2}}{sqrt{n-2}}$

Try

Using the facts:

• 
  $(1 + alpha/m)^m = e^alpha ( 1+ r_m)$, where $lim_{m to infty} sqrt{m}r_m = 0$
  


• 
  $Gamma(n+1) = n^{n + 1/2} e^{-n} sqrt{2 pi} (1 + r_n)$, where $lim_{n to infty} sqrt{n}r_n = 0$
  

Note :

$$
begin{aligned}
frac{Gamma((n-1)/2)}{Gamma(n/2)} &= frac{left(frac{n-3}{2}right)^{(n-2)/2} e^{-(n-3)/2}(sqrt{2pi}(1 + r_{1n})}{left(frac{n-2}{2}right)^{(n-1)/2} e^{-(n-2)/2}(sqrt{2pi}(1 + r_{2n})} \
&=left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right) times frac{sqrt{2}}{sqrt{n-2}}
end{aligned}
$$

But I'm stuck at how I should proceed to eliminate the $left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right)$ term.

A possible approach:

$$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}=tfrac{1}{sqrt{pi}},Bleft(tfrac{n-1}{2},tfrac{1}{2}right)=frac{1}{sqrt{pi}}int_{0}^{1}x^{frac{n-3}{2}}(1-x)^{-1/2},dx=frac{2}{sqrt{pi}}int_{0}^{1}frac{x^{n-2}}{sqrt{1-x^2}},dx $$
gives
$$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}= frac{2}{sqrt{pi}}int_{0}^{pi/2}left(costhetaright)^{n-2},dthetasimfrac{2}{sqrt{pi}}int_{0}^{+infty}expleft[-(n-2)frac{theta^2}{2}right]dtheta=sqrt{frac{2}{n-2}}. $$
Notice that the integral representation (as a moment) instantly gives that the LHS is a log-convex function. The asymptotic equivalence $sim$ can be seen as an instance of Laplace/Hayman's method.