Limit about Gamma function

How can we calculate the following ?:
limr→∞√r2 Γ((r−1)/2)Γ(r/2)=1 \lim_{r \to \infty}\,\sqrt{\,{r \over 2}\,}\,\
{\Gamma\left(\,\left(r – 1\right)/2\,\right) \over
\Gamma\left(\,r/2\,\right)} = 1

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Did you try using Stirling’s approximation?
– Stoke
2 days ago

  

 

@Stoke No… Can Stirling formula be applied to non-integers? But for advance, Thanks. If I use the formula, I can handle it.
– Heonjin Ha
2 days ago

  

 

The approximation is also good for the Gamma function. You can use it by converting back to factorial by Γ(n)=(n−1)!\Gamma(n)=(n-1)!
– Stoke
2 days ago

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4 Answers
4

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By Gautschi’s inequality with x+1=r2x+1=\frac{r}{2} and s=12s=\frac{1}{2},

√x≤Γ(x+1)Γ(x+s)≤√x+1\sqrt{x}\leq \frac{\Gamma(x+1)}{\Gamma(x+s)}\leq \sqrt{x+1} \tag{1}
the claim immediately follows by squeezing.

Hint

For this kind of problems, Stirling approximation is the key.

Consider A=√r2Γ(r−12)Γ(r2)⟹log(A)=12log(r)−12log(2)+log(Γ(r−12))−log(Γ(r2))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(Γ(m))=m(log(m)−1)+12(−log(m)+log(2π))+O(1m)\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 rr, you should find log(A)=34r+O(1r)\log(A)=\frac{3}{4 r}+O\left(\frac{1}{r}\right) and now, remembering that A=elog(A)A=e^{\log(A)} and Taylor again A=1+34r+O(1r)A=1+\frac{3}{4 r}+O\left(\frac{1}{r}\right) For illustration purposes, using r=100r=100, the exact value is ≈1.00758\approx 1.00758 while the simple asymptotics gives 1.007501.00750.

Hint. By Stirling’s formula (note that Γ(x+1)\Gamma(x+1) is an increasing function for x>0x>0 and Γ(x+1)=x!\Gamma(x+1)=x! if xx is a non-negative integer),
Γ(x+1)∼√2πx⋅xxex.\Gamma(x+1)\sim\sqrt{2\pi x}\cdot \frac{x^x}{e^x}.

\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\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}\,}\,}
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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}\bracks{{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{}} = \bbx{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}\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}\bracks{{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{}} = \bbx{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}\ds{\Gamma}-Duplication Formula to get rid of \ds{1/2}\ds{1/2}-factors. In this way, the \ds{\Gamma^{\, 2}\pars{r}}\ds{\Gamma^{\, 2}\pars{r}} function in the denominator is quite convenient.