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Difficult limit formula

  1. Dec 2, 2008 #1
    I have some reasons to believe that this equation is true:

    [tex]
    \lim_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \frac{(2n)!}{(n!)^2} = \frac{1}{\sqrt{\pi}}
    [/tex]

    Anyone having idea of the proof? I don't even know how to prove that the limit is strictly between zero and infinity.
     
  2. jcsd
  3. Dec 2, 2008 #2
    I can do it with Stirlings approximation.
     
  4. Dec 2, 2008 #3
    ok. Thank's for reminding of it.
     
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