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.
 A: 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 : 
$$\lim_{n\to \infty} {\frac{(2n-1)!!}{(2n)!!}} =  0$$
A: Since you are asking for an idea, then you can use property of double factorial. Click the given link, take a look equation $(9)$ and $(11)$. You will obtain
$$
\frac{(2n-1)!!}{(2n)!!}=\frac{(2n)!}{(2^n\ n!)^2}=\frac{(2n)!}{4^n(n!)^2},
$$
then use Stirling's approximation
$$
n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n
$$
as $n\to\infty$.
A: Using 
\begin{align}
(2n-1)!! &= \frac{2^{n} \Gamma(n+1/2)}{\sqrt{\pi}} \\
(2n)!! &= 2^{n} n!
\end{align}
then
\begin{align}
\frac{(2n-1)!!}{(2n)!!} = \frac{\Gamma(n+1/2)}{\sqrt{\pi} \Gamma(n+1)}.
\end{align}
Now using the duplication formula, $\Gamma(2x) = (2\pi)^{-1/2} 2^{2x-1/2} \Gamma(x) \Gamma(x+1/2)$, this fraction becomes
\begin{align}
\frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^{n}} \binom{2n}{n}.
\end{align}
Since $4^{n}$ grows faster than the binomial component it is then seen that
\begin{align}
\lim_{n \rightarrow \infty} \left[ \frac{(2n-1)!!}{(2n)!!} \right] = \lim_{n \rightarrow \infty} \left[ \frac{1}{4^{n}} \binom{2n}{n} \right] \rightarrow 0.
\end{align}
A: Consider $(2n)!$
$(2n)!$=$2n(2n-1)(2n-2).........2(1)$
     =$[(2n-1)(2n-3).....1][(2n)(2n-2)(2n-4).....2]$
     =$(2n-1)!!$$2^n$$n!$
which gives $(2n-1)!!=(2n)!$$/$$2^n$$n!$
$(2n)!!$$=$$(2n)(2n-2)........2$
    =$[2n][2n-2][2n-4]....2$
    =$2^n$$n!$

 so $(2n-1)!!$$/$$(2n)!!$$=$$(2n)!!$/$4^n$$[$(n)!$]^2$ 
$(2n-1)!!$$/$$(2n)!!$$=$$2n\choose n$$4^n$
using Stirling's Approximation
$2n\choose n$$=$$(2n/e)^{2n}\sqrt{2\pi*2n}$/$(n/e)^n\sqrt{2\pi*n}$
$=$$4^n$$[(n/e)^{2n}2\sqrt{\pi*n}]$/$[(n/e)^{2n}{2\pi*n}]$ 

$=$$4^n$/$\sqrt{\pi*n}$

which gives

$(2n-1)!!$$/$$(2n)!!$$ = $1/$\sqrt{\pi*n}$

finaly the limit $n$$\longrightarrow$$\infty$

$=$$0$
A: May this help a bit?
For positive integers, those relations hold:
$$(2n)!! = 2^n n!$$
$$(2n-1)!! = \frac{(2n)!}{2^n n!}$$
But I guess that your case need the general $n$ not only integers..
EDIT DUE TO THE COMMENTS BELOW
Using the relationships above and the Stirling approximations below, we find:
$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2^n n!)(2^n n!)} = \frac{(2n)!}{2^{2n} (n!)^2}$$
Applying the Stirling approximations:
$$\frac{(2n)!}{2^{2n} (n!)^2} = \frac{\sqrt{2\pi 2n} \left(\frac{2n}{e}\right)^{2n}}{2^{2n}(2\pi n)\left(\frac{n}{e}\right)^{2n}} = \frac{1}{\sqrt{\pi n}}$$
Chic goes to zero as $n$ goes to $+\infty$
A: $\lim\limits_{n\to \infty}\dfrac{(2n-1)!!}{(2n)!!}=0\quad$ and $\quad\lim\limits_{n\to \infty}\dfrac{(2n+1)!!}{(2n)!!}=\infty.~$ Where it really gets interesting, however, 
is when we attempt to evaluate their product. Their polar-opposite tendencies will cancel each 
other out, yielding $~\lim\limits_{n\to \infty}\dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac{(2n+1)!!}{(2n)!!}=\dfrac2\pi~,~$ which is the famous Wallis product. 
See also Basel problem for more information.
A: You can notice that $$\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 $0<a_k<1$.)
