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Proving a seqeunce converges

  1. Mar 22, 2008 #1
    [SOLVED] proving a seqeunce converges

    1. The problem statement, all variables and given/known data
    Prove that the sequence [itex]\frac{n!}{n^n}[/itex] converges to 0.


    2. Relevant equations



    3. The attempt at a solution
    Given [itex]\epsilon > 0[/itex], how do I find N? We know that the nth term is less than or equal to
    [tex]\left(\frac{n-1}{n}\right)^{n-1}[/tex]
    but that really does not help.
     
  2. jcsd
  3. Mar 22, 2008 #2

    tiny-tim

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    … in a fight, always pick off the smallest ones …

    Hi ehrenfest! :smile:

    Hint: write it (1/n)(2/n)(3/n)…(n/n);

    the early bits go down much faster than the later bits; so can you see a way of splitting off some of the early bits, and prove that they tend to zero? :smile:
     
  4. Mar 22, 2008 #3
    Yes, I was trying to do something like that. If I can show that any factor in that product goes to 0, then the whole thing has to because the rest will be less than 1. If I collect the first n/2 ceiling terms, then I get (1/2)^(n/2) if n is even, and that needs to got to zero, doesn't it? I see, thanks. When n is odd we get something similar.
     
  5. Mar 22, 2008 #4

    tiny-tim

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    That's right! :smile:

    (And don't forget, your proof should begin something like "For any epsilon, choose N such that 2^-N …")
     
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