How to prove that $\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0$ 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.

You can notice that $$\frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^n \frac{2k-1}{2k}
= \prod_{k=1}^n \left(1-\frac{1}{2k}\right).$$
So you want in fact calculate the infinite product
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1-\frac{1}{2k}\right) = \prod_{k=1}^\infty \left(1-\frac{1}{2k}\right)$$
Now you can use this fact: How to prove $\prod_{i=1}^{\infty} (1-a_n) = 0$ iff $\sum_{i=1}^{\infty} a_n = \infty$?
In fact, for this you do not need the equivalence, only one implication is sufficient: 
$$\prod_{k=1}^n (1-a_k) \le \frac1{\prod_{k=1}^n (1+a_k)} \le \frac1{1+\sum_{k=1}^n a_k}.$$
(We assume that $0<a_k<1$.)
