How to calculate the limit $\lim_{n\rightarrow \infty} \sqrt 2 \frac {\Gamma\left(\frac {n+1} 2\right)} {\Gamma\left(\frac {n} 2\right)}-\sqrt n$

Question

$$\lim_{n\rightarrow \infty} \sqrt 2 \frac {\Gamma\left(\frac {n+1} 2\right)} {\Gamma\left(\frac {n} 2\right)}-\sqrt n$$

It seems that using Stirling approximation doesn't work.

Answer 1

You do not need Stirling's approximation, but it has to work.

Let $r(n)=\Gamma((n+1)/2)/\Gamma(n/2)$. Since $$ \Gamma(\alpha)=\int_{0}^{+\infty} x^{\alpha}\frac{dx}{xe^x} $$ the $\Gamma$ function is log-convex on $\mathbb{R}^+$, hence $r(n)$ is an increasing function. We have $$ r(n)^2\leq r(n)r(n+1) = \frac{n}{2} \leq r(n+1)^2 $$ from which $\sqrt{2}\,r(n) = \sqrt{n}+o(1)$ can be easily deduced.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{n \to \infty}\braces{\root{2}\, {\Gamma\pars{\bracks{n + 1}/2} \over \Gamma\pars{n/2}} - \root{n}}} \\[5mm] = &\ \lim_{n \to \infty}\bracks{\root{2}\, {\pars{n/2 - 1/2}! \over \pars{n/2 - 1}!} - \root{n}} \\[5mm] = &\ \lim_{n \to \infty}\bracks{\root{2}\, {\root{2\pi}\pars{n/2 - 1/2}^{n/2}\expo{-n/2 + 1/2} \over \root{2\pi}\pars{n/2 - 1}^{n/2 - 1/2}\expo{-n/2 + 1}} - \root{n}} \\[5mm] = &\ \lim_{n \to \infty}\bracks{\root{2}\, {\pars{n/2}^{n/2}\,\pars{1 - 1/n}^{n/2} \over \pars{n/2}^{n/2 - 1/2}\pars{1 - 2/n}^{n/2 - 1/2}}\,\expo{-1/2} - \root{n}} \\[5mm] = &\ \lim_{n \to \infty}\pars{\root{2}\root{n \over 2} {\expo{-1/2} \over \expo{-1}}\,\expo{-1/2} - \root{n}} = \bbx{\large 0} \end{align}