TITLE: How to prove that $\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0$
QUESTION [7 upvotes]: So guys, how can I evaluate and prove that $$\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0.$$ Any ideas are welcomed.
$n!!$ is the double factorial, as explained in this  wolfram post.

REPLY [7 votes]: After some research and thinking I have realised the following :
We can observe that : $$(2K-1)*(2K+1) <  (2K)^2$$
Now, if we give values to K ( from 1 to n ) and the multiply all of them together we get :
$$ 1 * 3^2 * 5^2 * ... * (2n-1)^2 * (2n+1) < 2^2 * 4^2 * ... * (2n)^2 $$
Now by dividing with $2n+1$ and the right side plus taking the square root of the entire identiy we get : 
$$ {\frac {(2n-1)!!}{(2n)!!}} <  {\frac {1}{\sqrt{2n+1}}}$$
Since the second term has the limit 0 as n goes to infinity and the left side is positive we get that : 
$$\lim_{n\to \infty} {\frac{(2n-1)!!}{(2n)!!}} =  0$$