calculus
posted by Roy on .
Prove that the limit as n approaches infinity of ((n!)^2)/(2n)! is 0.
I'm thinking that you have to use the squeeze theorem, but I'm not quite sure how. help please?

how about something like this ....
n!/(2n)!
= n(n1)(n2)(n3) ... (2)(1)/ [(2n)(2n1)(2n2)(2n3) ..(2)(1) ]
= n(n1)(n2)(n3) ... (2)(1)/ 2n(2n1)(2)(n1)(2n3)(2)(n2)(2n5)(2)(n3) ... (2)(1) ]
notice that all of the factors of the numerator are found in the denominator, so ...
= 1/ [ 2^n(2n1)(2n3)(2n5) ... (2)(1) ]
clearly as n > ∞, the denominator  ∞
1/(very large) > 0
and the limit n!/(2n)! = 0