Limit about Gamma function

Question

How can we calculate the following ?: $$ \lim_{r \to \infty}\,\sqrt{\,{r \over 2}\,}\,\ {\Gamma\left(\,\left(r - 1\right)/2\,\right) \over \Gamma\left(\,r/2\,\right)} = 1 $$

Answer 1

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.

Answer 2

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 $1.00750$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.