How can I express the ration of double factorials $\frac{(2n+1)!!}{(2n)!!}$ as a single factorial?

Question

How can I change the double factorial of $$\frac{(2n+1)!!}{(2n)!!}$$ to single factorial?

Answer 1

$$(2n+1)!!=(2n+1)(2n-1)(2n-3) \cdots (3) (1)$$

$$=\frac{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4) \cdots (4)(3)(2) (1)}{(2n)(2n-2)(2n-4) \cdots (4)(2)}=\frac{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4) \cdots (4)(3)(2) (1)}{(2)(n)(2)(n-1)(2)(n-2) \cdots (2)(2)(2)(1)}$$

$$=\frac{(2n+1)!}{(2^n)n!}$$

Similarly,

$$(2n)!!=(2n)(2n-2)(2n-4) \cdots (4)(2)=(2)(n)(2)(n-1)(2)(n-2) \cdots (2)(2)(2)(1)=(2^n)n!$$

Answer 2

Hint 1: $(2n+1)!! = \frac{(2n+1)!}{(2n)!!}$

Hint 2: $(2n)!! = n! 2^n $.