Limit about Gamma function How can we calculate the following ?:
$$ \lim_{r \to \infty}\,\sqrt{\,{r \over 2}\,}\,\
{\Gamma\left(\,\left(r - 1\right)/2\,\right) \over
\Gamma\left(\,r/2\,\right)} = 1
$$
 A: By Gautschi's inequality with $x+1=\frac{r}{2}$ and $s=\frac{1}{2}$,
$$\sqrt{x}\leq \frac{\Gamma(x+1)}{\Gamma(x+s)}\leq \sqrt{x+1} \tag{1}$$
the claim immediately follows by squeezing.
A: Hint
For this kind of problems, Stirling approximation is the key.
Consider $$A=\sqrt\frac{r}{2} \frac{\Gamma(\frac{r-1}{2})}{\Gamma(\frac{r}{2})}\implies \log(A)=\frac 12 \log(r)-\frac 12 \log(2)+\log\left(\Gamma(\frac{r-1}{2})\right)-\log\left(\Gamma(\frac{r}{2})\right)$$ Stirling approximation write $$\log\left(\Gamma(m)\right)=m (\log (m)-1)+\frac{1}{2} \left(-\log (m)+\log (2 \pi
   )\right)+O\left(\frac{1}{m}\right)$$ Apply to each factorial and simplify.
If you use it and continue with Taylor series for infinitely large values of $r$, you should find $$\log(A)=\frac{3}{4 r}+O\left(\frac{1}{r}\right)$$ and now, remembering that $A=e^{\log(A)}$ and Taylor again $$A=1+\frac{3}{4 r}+O\left(\frac{1}{r}\right)$$ For illustration purposes, using $r=100$, the exact value is $\approx 1.00758$ while the simple asymptotics gives $1.00750$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\begin{align}
&\lim_{r \to \infty}\braces{\root{r \over 2}\,
{\Gamma\pars{\bracks{r - 1}/2} \over \Gamma\pars{r/2}}} =
\lim_{r \to \infty}\bracks{\root{r + 1}\,
{\Gamma\pars{r + 1/2} \over \Gamma\pars{r + 1}}}\qquad
\pars{~{r \over 2}\ \mapsto\ r + 1~}
\\[5mm] = &\
\root{2\pi}\lim_{r \to \infty}\bracks{{2^{1/2 - 2r} \over \root{r + 1}}\,
{\Gamma\pars{2r} \over \Gamma^{\, 2}\pars{r}}}\qquad\qquad\qquad
\pars{~\Gamma\mbox{-}Duplication\ Formula\ \mbox{and}\ Recurrence~}
\\[5mm] = &\
\root{2\pi}\lim_{r \to \infty}\braces{{2^{1/2 - 2r} \over \root{r + 1}}\,
{\root{2\pi}\pars{2r - 1}^{2r - 1/2}\expo{-2r + 1} \over
\bracks{\root{2\pi}\pars{r - 1}^{r - 1/2}\expo{-r + 1}}^{\, 2}}}
\quad\pars{~Stirling\ Asymptotic\ Expansion~}
\\[5mm] = &\
\lim_{r \to \infty}\bracks{{2^{1/2 - 2r} \over \root{r + 1}}\,
{\pars{2r - 1}^{2r - 1/2}\expo{-2r + 1} \over
\pars{r - 1}^{2r - 1}\expo{-2r + 2}}} =
\lim_{r \to \infty}\braces{{r^{1/2} \over \root{r + 1}}\,
{\bracks{1 - \pars{1/2}/r}^{2r} \over\pars{1 - 1/r}^{2r}\expo{}}}
\\[5mm] & =
{\pars{\expo{-1/2}}^{2} \over \pars{\expo{-1}}^{2}\expo{}} =
\bbox[#ffd,10px,border:1px dotted navy]{1}
\qquad\qquad\qquad\qquad\qquad\qquad
\pars{~\mbox{Note that}\ \lim_{n \to \infty}\pars{1 + {x \over n}}^{n} = \expo{x}~}
\end{align}

We used the $\ds{\Gamma}$-Duplication Formula to get rid of $\ds{1/2}$-factors. In this way, the $\ds{\Gamma^{\, 2}\pars{r}}$ function in the denominator is quite convenient.

