<s>[INST] Demonstrate that $\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} \simeq 1.7 \sqrt{n}$ As in the title, I know that
$\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} = \frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1} \simeq 1.7 \sqrt{n}$
Could you give some hint to prove it?
(should I look the series expansion of $\sqrt{n}$?)
Thank you anyway!
 [/INST] Use the Stirling formula:
$$\frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1}
=\frac{(2n - 2)^2(2n - 4)^2\cdots 4^2 \cdot 2^2}
{(2n - 2)!}
\\=
2^{2n-2}
\frac{(n - 1)^2(n - 2)^2\cdots 2^2 \cdot 1^2}
{(2n - 2)!}
\\=2^{2n-2}
\frac{((n - 1)!)^2}{(2n - 2)!}
\sim 2^{2n-2}\frac{(n-1)^{2n-2}e^{-2n+2}2\pi n}
{(2n - 2)^{2n-2}e^{-2n+2}\sqrt{2\pi 2n}}=
{\sqrt{\pi n}}
$$
</s>