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!

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Use Stirling's formula. It doesn't diverge; in fact, since $4^n = 2^{2n} = \sum {2n \choose k}$ it is bounded above by $1$. –  Qiaochu Yuan Dec 1 '12 at 23:22
–  Raymond Manzoni Dec 1 '12 at 23:33

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 $\lim_{n\to \infty}a_n = 0$.

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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 $n \rightarrow \infty$.

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