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!

Also with Stirling's approximations, the comparison of numerical results :

