Inequality $(9)$ in this answer says that
$$
\frac{4^n}{\sqrt{\pi\!\left(n+\frac14+\frac1{16n+12}\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}\tag1
$$
Therefore,
$$
\begin{align}
\frac{\color{#C00}{(2n)!!}}{\color{#090}{(2n-1)!!}}
&=\frac{\color{#C00}{2^nn!}\color{#090}{2^nn!}}{\color{#090}{(2n)!}}\tag{2a}\\[3pt]
&=\frac{4^n}{\binom{2n}{n}}\tag{2b}\\
&\in\left[\sqrt{\pi\!\left(n+\tfrac14\right)},\sqrt{\pi\!\left(n+\tfrac14+\tfrac1{16n+12}\right)}\right]\tag{2c}
\end{align}
$$
SquaringExplanation:
$\text{(2a):}$ rewrite double factorials in terms of factorials
$\text{(2b):}$ $\binom{2n}{n}=\frac{(2n)!}{n!n!}$
$\text{(2c):}$ apply $(1)$
Squaring $(2)$ and subtracting $\pi n$ gives $$ \frac\pi4\le\left(\frac{(2n)!!}{(2n-1)!!}\right)^2-\pi n\le\frac\pi4+\frac\pi{16n+12}\tag3 $$ Evaluate the limit with the Squeeze Theorem.
