How would I solve the sum n=1 to infinity of (n!)^2/(2n)!

Question

anonymous · Answer

$$\sum_{n=1}^{\infty} \frac{ (n!)^2 }{ (2n)! }$$

anonymous · Answer

Actually I take that back, I don't need to solve, just to establish convergence or divergence

anonymous · Answer

might need someone else to check what i did

since its from 1 to infinity you can just plug in a really big number into the equation
so the top gets really big and the bottom too 
but the bottom is multiplied by 2 so it is twice as large than the top, since denominator is bigger than numerator, and if you plug in a huge number it will get close to 0, so your answer is 0

anonymous · Answer

i just learned this and had a test on it, i should know... heres a list of tests http://www.math.com/tables/expansion/tests.htm

anonymous · Answer

i think this test will be the best:
Limit Comparison Test
If lim (n-->) (an / bn) = L,
where an, bn > 0 and L is finite and positive,
then the series  an and  bn either both converge or both diverge.