calculus

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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?

  • calculus -

    how about something like this ....

    n!/(2n)!
    = n(n-1)(n-2)(n-3) ... (2)(1)/ [(2n)(2n-1)(2n-2)(2n-3) ..(2)(1) ]
    = n(n-1)(n-2)(n-3) ... (2)(1)/ 2n(2n-1)(2)(n-1)(2n-3)(2)(n-2)(2n-5)(2)(n-3) ... (2)(1) ]
    notice that all of the factors of the numerator are found in the denominator, so ...

    = 1/ [ 2^n(2n-1)(2n-3)(2n-5) ... (2)(1) ]

    clearly as n ---> ∞, the denominator --- ∞
    1/(very large) ---> 0

    and the limit n!/(2n)! = 0

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