Limit of equation regarding ratio of gamma function

Question

I get the result from Wolfram that $$\lim_{a\to \infty}a-\frac{\Gamma(a+1/2)^2}{\Gamma(a)^2}=\frac{1}{4}.$$

I am trying to prove it. It seems the Stirling's formula cannot be used here. Can anyone help me on this?

Thank you very much!

I tried to use Stirling's approximation as:

\begin{equation} \begin{aligned} a-\frac{\Gamma(a+1/2)^2}{\Gamma(a)^2} &\to a-\frac{(a-1/2)^{2a}}{(a-1)^{2a-1}}e^{-1}\\ &=a-\big(\frac{a-1+1/2}{(a-1)}\big)^{2(a-1)}e^{-1}\frac{a^2-a+1/4}{a-1}\\ &\to a-a-\frac{1}{4(a-1)} \end{aligned} \end{equation}

Answer 1

Considering Stirling approximation $$\log (\Gamma (p))=p (\log (p)-1)-\frac 12 \log(p)+\frac{1}{2} \log (2 \pi )+\frac{1}{12 p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ apply it to $$A=\frac{\Gamma \left(a+\frac{1}{2}\right)}{\Gamma (a)}\implies \log(A)=\log \left(\Gamma \left(a+\frac{1}{2}\right)\right)-\log (\Gamma (a))$$ and continue with Taylor expansion for large values of $a$ to get $$\log(A)=\frac{1}{2} \log \left({a}\right)-\frac{1}{8 a}+\frac{1}{192 a^3}+O\left(\frac{1}{a^4}\right)$$ $$\log(A^2)=2\log(A)= \log \left({a}\right)-\frac{1}{4 a}+\frac{1}{96 a^3}+O\left(\frac{1}{a^4}\right)$$ Continue with Taylor, using $A^2=e^{\log(A^2)}$ to get $$A^2=a-\frac{1}{4}+\frac{1}{32 a}+\frac{1}{128 a^2}+O\left(\frac{1}{a^3}\right)$$ which makes $$a-\left(\frac{\Gamma \left(a+\frac{1}{2}\right)}{\Gamma (a)} \right)^2=\frac{1}{4}-\frac{1}{32 a}-\frac{1}{128 a^2}+O\left(\frac{1}{a^3}\right)$$ which shows the limit and how it is approached.

Moreover, it is a quite good approxiamtion of the function even for small values of $a$ as shown in the table $$\left( \begin{array}{ccc} a & \text{approximation} & \text{exact} \\ 1 & 0.210938 & 0.214602 \\ 2 & 0.232422 & 0.232854 \\ 3 & 0.238715 & 0.238835 \\ 4 & 0.241699 & 0.241747 \\ 5 & 0.243438 & 0.243461 \\ 6 & 0.244575 & 0.244588 \\ 7 & 0.245376 & 0.245384 \\ 8 & 0.245972 & 0.245977 \\ 9 & 0.246431 & 0.246435 \\ 10 & 0.246797 & 0.246800 \end{array} \right)$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With A & S Table $\mathbf{\color{black}{6.1.47}}$ identity:

\begin{align} a^{-1/2}\,{\Gamma\pars{a + 1/2} \over \Gamma\pars{a}} & \,\,\,\stackrel{\mrm{as}\ a\ \to\ \infty}{\sim}\,\,\, 1 - {1 \over 8a} + {1 \over 128a^{2}} \\[5mm] \implies \bracks{\Gamma\pars{a + 1/2} \over \Gamma\pars{a}}^{2} & \,\,\,\stackrel{\mrm{as}\ a\ \to\ \infty}{\sim}\,\,\, a\pars{1 - {1 \over 8a} + {1 \over 128a^{2}}}^{2} \,\,\,\stackrel{\mrm{as}\ a\ \to\ \infty}{\sim}\,\,\, \bbx{a - \color{red}{1 \over 4} + {1 \over 32a}} \end{align}

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$$ \bbx{a - \bracks{\Gamma\pars{a + 1/2} \over \Gamma\pars{a}}^{2} \,\,\,\stackrel{\mrm{as}\ a\ \to\ \infty}{\sim}\,\,\, \color{red}{1 \over 4} - {1 \over 32a}} $$

Answer 3

$\Gamma$ is log-convex by the Bohr-Mollerup theorem/characterization, hence it is enough to prove $$ \lim_{n\to +\infty} n-\left(\frac{n\sqrt{\pi}}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{4}.\tag{1} $$ On the other hand $$ \frac{1}{4^n}\binom{2n}{n} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),\qquad \left(\frac{1}{4^n}\binom{2n}{n}\right)^2=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{2}$$ and $$\pi\left(\frac{n}{4^n}\binom{2n}{n}\right)^2=n\prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)^{-1}\tag{3}$$ by Wallis' product. We also have $$\prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)^{-1}=\exp\sum_{k>n}\left[-\frac{1}{4k(k-1)}+O\left(\frac{1}{k^4}\right)\right]=\exp\left(-\frac{1}{4n}\right)\left(1+O\left(\frac{1}{n^3}\right)\right) $$ and the RHS of the last line equals $$\left(1-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\right)\left(1+O\left(\frac{1}{n^3}\right)\right)=1-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\tag{4}$$ so the claim is proved without resorting to the full power of Stirling's approximation/inequality.