How to show $\frac{\Gamma((n-1)/2)}{\Gamma(n/2)} \approx \frac{\sqrt{2}}{\sqrt{n-2}}$ 
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 :
$$
\begin{aligned}
\frac{\Gamma((n-1)/2)}{\Gamma(n/2)} &= \frac{\left(\frac{n-3}{2}\right)^{(n-2)/2} e^{-(n-3)/2}(\sqrt{2\pi}(1 + r_{1n})}{\left(\frac{n-2}{2}\right)^{(n-1)/2} e^{-(n-2)/2}(\sqrt{2\pi}(1 + r_{2n})} \\
&=\left( \left(\frac{n-3}{n-2}\right)^{(n-2)/2} \sqrt{e} \frac{1+r_{1n}}{1 + r_{2n}} \right)  \times \frac{\sqrt{2}}{\sqrt{n-2}} 
\end{aligned}
$$
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:
$$ \frac{\Gamma\left(\tfrac{n-1}{2}\right)}{\Gamma\left(\tfrac{n}{2}\right)}=\tfrac{1}{\sqrt{\pi}}\,B\left(\tfrac{n-1}{2},\tfrac{1}{2}\right)=\frac{1}{\sqrt{\pi}}\int_{0}^{1}x^{\frac{n-3}{2}}(1-x)^{-1/2}\,dx=\frac{2}{\sqrt{\pi}}\int_{0}^{1}\frac{x^{n-2}}{\sqrt{1-x^2}}\,dx $$
gives
$$ \frac{\Gamma\left(\tfrac{n-1}{2}\right)}{\Gamma\left(\tfrac{n}{2}\right)}= \frac{2}{\sqrt{\pi}}\int_{0}^{\pi/2}\left(\cos\theta\right)^{n-2}\,d\theta\sim\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\exp\left[-(n-2)\frac{\theta^2}{2}\right]d\theta=\sqrt{\frac{2}{n-2}}. $$
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.
