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:

 A: $$f(n)=\sqrt{2}\,\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} - \sqrt{n}=\sqrt{2}\,y- \sqrt{n}$$
Take logarithms
$$\log(y)=\log \left(\frac{\Gamma \left(\frac{n+1}{2}\right)}{\Gamma
   \left(\frac{n}{2}\right)}\right)=\log \left(\Gamma \left(\frac{n+1}{2}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}\right)\right)$$ Use Stirling approximation twice
$$\log(y)=\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{1}{4 n}+\frac{1}{24
   n^3}+O\left(\frac{1}{n^5}\right)$$ Continue with Taylor
$$y=e^{\log(y)}=\sqrt{\frac n 2}\Bigg[1-\frac{1}{4 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$
$$f(n)=-\frac 1{4\sqrt {n}}\Bigg[1-\frac{1}{8 n}-\frac{5}{32 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$
Use it for $n=9$; rhe above truncated series gives
$$f(9) \sim -\frac{2551}{31104}=-0.0820152\cdots$$ while the exact value is
$$f(9)=-3+\frac {128}{35} \sqrt{\frac 2 \pi}=-0.0820222\cdots$$
A: 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 $0$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{\root{2}{\Gamma\pars{n/2 + 1/2} \over \Gamma\pars{n/2}} - \root{n}}} =
\lim_{n \to \infty}\bracks{\root{2}{\pars{n/2 - 1/2}! \over \pars{n/2 - 1}!} - \root{n}}
\\[5mm] = & \
\lim_{n \to \infty}\bracks{\root{2}{\root{2\pi}\pars{n/2 - 1/2}^{n/2}\expo{-n/2 + 1/2} \over
\root{2\pi}\pars{n/2 - 1}^{n/2 - 1/2}\,\expo{-n/2+1}} - \root{n}}
\ \substack{Stirling\ Asymptotic\\[1mm] Behavior}  \\[5mm] = & \
\lim_{n \to \infty}\bracks{\root{2}{\pars{n/2}^{n/2}\,\pars{1 - 1/n}^{n/2} \over
\pars{n/2}^{n/2 - 1/2}\,\,\,\pars{1 - 2/n}^{n/2}}\expo{-1/2} - \root{n}}
\\[5mm] = & \
\lim_{n \to \infty}\bracks{\root{2}{\expo{-1/2} \over
\root{2/n}\expo{-1}}\expo{-1/2} - \root{n}} = {\large 0}
\end{align}
