I have redone this problem many times, and can't seem to figure it out!

The problem is to use the ratio test to find the limit and convergence of this series:

$\displaystyle \sum_{i=1}^\infty \frac{(2n)!}{n!n+3)!}$

Now, using the ratio test, I come up with this:

$\displaystyle \lim_{n \rightarrow \infty} \frac{n!(n+3)!(2(n+1))!}{(2n)!(n+1)!(n+4)!}$

How do I expand this to cancel the series out? This is what I tried:

$\displaystyle \lim_{n \rightarrow \infty} \frac{(1*2...n)(1*2...(n+1)(n+2)(n+3)) (24*720...(2(n+1)) )}{(2*24*720...(2n)) (1*2...(n+1)) (1*2...(n+1)(n+2)(n+3)(n+4))}$

However, when I cancel everything out, I get $\displaystyle \lim_{n \rightarrow \infty} \frac{2n}{4n(n+4)}$ which results in the limit being 0.

But 0 is not the correct limit.

Any help would be very appreciated!