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Stirling's approximation limit problem

  1. May 26, 2009 #1
    Hi, I don't undestand this limits to infinity.

    [ (2n)! / (n!)^2 ]^1/n


    [ ( n^n + 2^n ) / (n! + 3^n) ]^1/n

    I've absolutely no idea how the first one can be "4", and the seconda one "e". Assuming (1 + 1/n)^n = e I don't see how can I go from that form to this one. Any hint is appreciated.
  2. jcsd
  3. May 26, 2009 #2


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    Welcome to PF!

    Hi nitai108! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Hint: what, approximately, is (2n)! / n! ?
    Hint: which bits of the top and the bottom can you ignore? :wink:
  4. May 26, 2009 #3
    Re: Limits

    To evaluate the first limit I would use Stirling's approximation, namely that [tex] n! \approx n^{n}e^{-n} [/tex] (and [tex] (2n)! \approx (2n)^{2n}e^{-2n} [/tex]) for large n.
    Last edited: May 27, 2009
  5. May 27, 2009 #4
    Re: Limits

    For the second limit, do what tiny-tim suggested and then use Stirling's approximation.
  6. May 28, 2009 #5
    Re: Limits

    Thanks tiny-tim, and thanks Random Variable, I did solve them pretty easily with the Stirling's approximation.

    Is there any other way to solve them, without Stirling's approx.?
  7. May 28, 2009 #6


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    The first one, you could say (2n)! is approximately (2n)2(2n-2)2 … 22

    (but you'd have to prove that rigorously)

    … which makes it fairly easy :wink:
  8. May 29, 2009 #7
    Re: Limits

    Ok, thanks a lot.
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