Evaluating $\lim_{n \to \infty}\left[\,\sqrt{\,2\,}\,\frac{\Gamma\left(n/2 + 1/2\right)}{\Gamma\left(n/2\right)} - \,\sqrt{\,n\,}\right]$ 
I am trying to find this limit:
$$\lim_{n \to \infty}\left[\,\sqrt{\,2\,}\,\frac{\Gamma\left(n/2 + 1/2\right)}{\Gamma\left(n/2\right)} - \,\sqrt{\,n\,}\right]$$

I know that
$$\frac{\sqrt{2}\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})}=\begin{cases}\sqrt{\frac{\pi}{2}} \frac{(n-1)!!}{(n-2)!!},\ n\ is\ even\sim \sqrt{n-1}\\ \sqrt{\frac{2}{\pi}} \frac{(n-1)!!}{(n-2)!!},\ n\ is\ odd \sim\sqrt{n+1}\end{cases}$$
However, I can not find the limit above.
Based on the numeric study, it seems like the limit is 0:


Using the first few terms of http://dlmf.nist.gov/5.11.E13 with $a=1/2$, $b=0$ and $z=n/2$, we find
$$
\frac{{\Gamma (n/2 + 1/2)}}{{\Gamma (n/2)}} = \sqrt {\frac{n}{2}} \left( {1 - \frac{1}{{4n}} + \mathcal{O}\!\left( {\frac{1}{{n^{2 } }}} \right)} \right).
$$
Thus,
$$
\sqrt 2 \frac{{\Gamma (n/2 + 1/2)}}{{\Gamma (n/2)}} - \sqrt n  =  - \frac{1}{{4\sqrt n }} + \mathcal{O}\!\left( {\frac{1}{{n^{3/2} }}} \right)
$$
and your limit is indeed $0$.
