$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}=\prod_{k=1}^{n}\bigl(1-\frac{1}{2k}\bigr)$ 
i cant see why we have :

*

*$$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$$


*$$\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$$

Even i see the notion of Double factorial
this question is related to that one : Behaviour of the sequence $u_n = \frac{\sqrt{n}}{4^n}\binom{2n}{n}$

*

*For  $\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$
$\dfrac{(2n)!}{4^n n!^2}=\dfrac{(2n)!}{2^{2n} n!^2}=\dfrac{(2n)\times (2n-1)!}{2^{2n} (n\times (n-2)!)^2}$

*

*For  $\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$
note that $n!! = \prod_{i=0}^k (n-2i) = n (n-2) (n-4) \cdots$
$\dfrac{(2n-1)!!}{(2n)!!}=\dfrac{\prod_{i=0}^k (2n-1-2i)}{\prod_{i=0}^k (2n-2i)}$
 A: Note first that $(2n)!=(2n)!!(2n-1)!!$, because $(2n)!!$ gives you the even factors in $(2n)!$, and $(2n-1)!!$ gives you the odd factors. Now
$$\begin{align*}
(2n)!!&=(2n)(2n-2)(2n-4)\ldots(4)(2)\\
&=\big(2n\big)\big(2(n-1)\big)\big(2(n-2)\big)\ldots\big(2(2)\big)\big(2(1)\big)\\
&=2^nn!\;,
\end{align*}$$
so
$$\frac{(2n)!}{4^nn!^2}=\frac{(2n)!!(2n-1)!!}{2^{2n}n!^2}=\frac{2^nn!(2n-1)!!}{2^{2n}n!^2}=\frac{(2n-1)!!}{2^nn!}=\frac{(2n-1)!!}{(2n)!!}\;.$$
Now 
$$\begin{align*}
\frac{(2n-1)!!}{(2n)!!}&=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdot\ldots\cdot\frac34\cdot\frac12\\\\
&=\left(1-\frac1{2n}\right)\left(1-\frac1{2n-2}\right)\left(1-\frac1{2n-4}\right)\ldots\left(1-\frac14\right)\left(1-\frac12\right)\\\\
&=\prod_{k=1}^n\left(1-\frac1{2k}\right)\;.
\end{align*}$$
A: $\dfrac{(2n)!}{4^nn!^2}=\dfrac{(2n)!!(2n-1)!!}{(2^nn!)^2}=\dfrac{(2n)!!(2n-1)!!}{((2n)!!)^2}=\dfrac{(2n-1)!!}{(2n)!!}$
$\prod_{k=1}^n(1-\frac{1}{2k})=\prod_{k=1}^n\frac{2k-1}{2k}=\dfrac{\prod_{k=1}^n(2k-1)}{\prod_{k=1}^n(2k)}=\frac{(2n-1)!!}{(2n)!!}$
