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Homework Help: Factorial and limits

  1. Feb 7, 2010 #1
    So I've been asked to prove that:

    lim (n-->infinity) [2^n]/sqrt(n!) = 0

    I've tried fiddling with Stirling and L'Hospital, but can't find my way through it.

    Any thoughts?
  2. jcsd
  3. Feb 7, 2010 #2


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    Show us some of your 'fiddling'. If you use Stirling, it's not that hard.
  4. Feb 7, 2010 #3
    Stirling brought me to:

    lim n-->inf [2^n * e^(n/2)]/n^(n/2+1/4)

    I've also been able to express it as:

    lim n-->inf [2 * e^(1/2)]^n / n^(n/2+1/4)

    And I won't show you L'Hospital's results since it gets horrible... Anyways, I feel there's something obvious about it that I cannot see. L'Hospital won't help me as long as I have something^n in my numerator...

    I'm turning in circle.
  5. Feb 7, 2010 #4


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    Look at the log of your expression. log(2^n/sqrt(n!))=log(2^n)-log(sqrt(n!))~n*log(2)-(1/2)*n*log(n). That goes to -infinity as n->infinity, right? That means 2^n/sqrt(n!) goes to zero, also right? Notice I threw out all terms in Stirling's approximation except for log(n!)~n*log(n). The wouldn't be important unless the n*log(n) had cancelled out. It grows faster than all the other terms.
    Last edited: Feb 7, 2010
  6. Feb 8, 2010 #5
    Thanks for the hint, but I think I may have found another way to prove it.

    2^n may be splitted into n-1 products of n: 2 * 2 * 2 * 2 * 2 * ... * 2 <-- doing this n times
    sqrt(n!) may also be splitted into products like: sqrt(1) * sqrt(2) * sqrt(3) * ... * sqrt(n)

    Thus, I may recreate products of fraction like:
    [2/sqrt(1)] * [2/sqrt(2)] * [2/sqrt(3)] * [2/sqrt(4)] * ... [2/sqrt(n)]

    This makes it obvious that the limit converges to 0 since denominators increase, but numerators stay constant. Right?
  7. Feb 8, 2010 #6


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    Sure. That's another way. To phrase it differently, if f(n)=n!/sqrt(n!) then f(n+1)/f(n)=2/sqrt(n+1). That ratio is less than 1/2 for large n. You can appeal to sort of a ratio test.
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