Prove $(2.n+1)!!=\frac{(2.n+1)!}{2^n.n!}$ using Gamma Function

Question

Prove $(2.n+1)!!=\frac{(2.n+1)!}{2^n.n!}$ using Gamma Function.

$\Gamma(1+z).\Gamma(z+1/2)=2^{-2.z}.\sqrt{\pi}.\Gamma(2.z+1)$

$\Gamma(z+1/2)=\sqrt{\pi}.2^{-z}.(2.z-1)!!$

and

$\Gamma(1+z)=(2.z)!!.2^{-n}$

I think that at least one of these expressions is useful for this problem but I don't know which.

Answer 1

You can do this verbally without any equations.

The double factorial is the product of the odd numbers from one through 2n+1. What is omitted are the even numbers from 2 to 2n. There are n of these, so their product is $2^nn!$. And this is what you want.

I don't see how the gamma function comes into this at all unless you define the double factorial in terms of it.