Demonstrate that $\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} \simeq 1.7 \sqrt{n}$

Question

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!

Answer 1

Use the Stirling formula:

$$\frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1} =\frac{(2n - 2)^2(2n - 4)^2\cdots 4^2 \cdot 2^2} {(2n - 2)!} \\= 2^{2n-2} \frac{(n - 1)^2(n - 2)^2\cdots 2^2 \cdot 1^2} {(2n - 2)!} \\=2^{2n-2} \frac{((n - 1)!)^2}{(2n - 2)!} \sim 2^{2n-2}\frac{(n-1)^{2n-2}e^{-2n+2}2\pi n} {(2n - 2)^{2n-2}e^{-2n+2}\sqrt{2\pi 2n}}= {\sqrt{\pi n}} $$

Answer 2

Hint Take the square root of the Wallis product...

Answer 3

Just extending what mookid gave, you can have an extremely good approximation using $$ \frac{(2n - 2)!!}{(2n - 3)!!} \simeq \sqrt{\pi n } \left(1-\frac{3}{8 n}-\frac{7}{128 n^2}\right)$$

Answer 4

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