limit of $\frac{(2n)!}{4^n(n!)^2}$ I'd love to understand the behaviour of the sequence 
$$
\frac{(2n)!}{4^n(n!)^2} \text{as } n \to \infty
$$
the first step would be to simplify this to
$$
\frac{(2n)(2n-1)(2n-2)\cdots(n+1)}{4^n \cdot n(n-1)(n-2)\cdots 2} 
$$
and then factor out $2$ to get
$$
\frac{1}{2^n}\cdot\frac{(n)(n-1/2)(n-1)\cdots(n - (n-1)/2)}{n(n-1)(n-2)\cdots 2} 
$$
if I can now get the second term to be strictly larger than $2^n$ then I would be done - but how can I do this ? thanks so much for help!!
P.S. this not a HW question - though it grew out of one where I had to find the radius of convergence for a power series - this is the series evaluated at the end points. If I can show that the above sequence does not converge to $0$ then I know that the power series diverges at the endpoints, this is what I'd love to find out!
 A: Rewrite the sequence  as $$ a_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n) }. $$
Show that $$a_n \le \frac{1}{\sqrt{3n+1}}$$ using induction.
Conclude by Sandwich theorem that $\lim_{n\to \infty}a_n = 0$.
A: As mentioned before, you can rewrite the sequence as:
$$ a_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n) } $$
The following inequalities:
$1\cdot3<2^{2}$
$3\cdot5<4^{2}$
$5\cdot7<6^{2}$
$\cdots\cdots\cdots\cdots$
$(2n-3)(2n-1)<(2n-2)^{2}$
$(2n-1)(2n+1)<\left(  2n\right)  ^{2}$
multiplied lead to $a_{n}^{2}\cdot(2n+1)<1$, which means that $a_{n}<\dfrac
{1}{\sqrt{2n+1}}$. This is enough to guarentee that the limit of $a_{n}$ is
$0$.
A: Using  $2^n n! = 2 \cdot 4 \cdots  (2n)$ you can rewrite the sequence as
$$ a_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n) } $$
It is rather well known that 
$$\frac{1 / a_n} { \sqrt{n} } = \frac{\frac{2 \cdot 4 \cdots (2n) }{1 \cdot 3 \cdots (2n-1)}}{\sqrt n }  \rightarrow \sqrt \pi $$
So by taking the inverse of this, $a_n \rightarrow 0$ for $n \rightarrow \infty$.
