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Proof that $(n!)^2/(2n)!$ converges to $0$. I take following steps:

  1. $(n!)^2/(2n)(2n-1)\cdots(n!) = (n!)/(2n)(2n-1)\cdots(n-1)$. I assume (do I need to prove?) that $n!$ divides $(2n)(2n-1)\cdots(n-1)$.
  2. So I have at the end $1/K$ ($K$ is the remainder after division of the denominator by $n!$).
  3. $1/K$ as increases with increasing $n$ converges to $0$, is a null sequence.

Thanks for any advice.

  • 0
    $K$ isn't a remainder, it is the quotient. You haven't shown that $K$ increases as $n$ increases, much less that $K$ has infinite limit, so you haven't proven anything yet.2012-04-23
  • 0
    Where you have $n-1$, you actually get $n+1$.2012-04-23
  • 0
    I would say that you do need to prove that $n!$ divides $(2n)!/(n!)$. You also need to at least provide some sort of lower-bound of $K$ as a function of $n$.2012-04-23
  • 0
    K increases.Consider n+1. (2(n+1))!/(n+1)!^2. Gives us (2n+2)*(2n+1)*(2n)....(n-1) divided by (n+1)! (n+1)! already was divised by stopping at n-1.2012-04-24
  • 0
    consider n+1 then (2(n+1))!/(n+1)!^2. We get (2n+2)*(2n+1)/(n+1) when we divide twice by (n+1)!. This increases K surely. 1/K converges to 0.2012-04-24

3 Answers 3