Limit of equation regarding ratio of gamma function 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}
 A: $\Gamma$ 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

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