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!