A text I am reading claims that
$$(-g)^{n} \frac{(4 n-1) ! !}{n !} \stackrel{n \gg 1}{\sim}\left(-\frac{g n}{e}\right)^{n}$$
I know the stirling approximation $n!\stackrel{n \gg 1}{\sim} n^n e^{-n}$. But how do I go about approximating the double factorial in the numerator?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\pars{-g}^{n}\,{\pars{4n - 1}!! \over n!}} = {\pars{-g}^{n} \over n!}\,\prod_{k = 0}^{2n - 1}\pars{2k + 1} = {\pars{-g}^{n} \over n!}\,2^{2n}\prod_{k = 0}^{2n - 1}\pars{k + {1 \over 2}} \\[5mm] = &\ {\pars{-g}^{n} \over n!}\,2^{2n}\,\pars{1 \over 2}^{\overline{2n}} = {\pars{-g}^{n} \over n!}\,2^{2n}\,{\Gamma\pars{1/2 + 2n} \over \Gamma\pars{1/2}} = {2^{2n}\,\pars{-g}^{n} \over \root{\pi}}\,{\pars{2n - 1/2}! \over n!} \\[5mm] &\ \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {2^{2n}\,\pars{-g}^{n} \over \root{\pi}}\, {\root{2\pi}\pars{2n - 1/2}^{2n}\expo{-2n + 1/2} \over \root{2\pi}n^{n + 1/2}\,\, \expo{-n}} \\[5mm] = &\ {2^{2n}\,\pars{-g}^{n} \over \root{\pi}}\, {2^{2n}n^{2n}\,\, \bracks{1 - \pars{1/2}/\pars{2n}}^{\,2n}\expo{1/2} \over n^{n + 1/2}}\,\expo{-n} \\[5mm] &\ \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {2^{4n}\,\pars{-g}^{n} \over \root{\pi}}\, n^{n - 1/2}\,\expo{-n} = \bbx{{1 \over \root{\pi}}\pars{-16gn \over \expo{}}^{n}\,{1 \over n^{1/2}}} \\ & \end{align}