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[Calculus] Sequence Limits: n -> infinity (n!/n^n)(Use Sandwich Rule?)

  1. May 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Use sandwich Rule to find the limit lim n> infinity (a_n) of the sequences, for which the nth term, a_n, is given.


    2. Relevant equations
    [itex] ^{lim}_{n\rightarrow∞}\frac{n!}{n^{n}}[/itex]


    3. The attempt at a solution
    I know by just looking at it, n^n Approaches infinity much faster than n! which results in limit approaching 0, which is the answer. But the question says to use Sandwich Rule? I don't know which 2 functions to use to bound n!/n^n between? Usually if there was a sin function, I could start with it between -1 and 1. But I don't know where to start for this question?
     
  2. jcsd
  3. May 3, 2013 #2

    Dick

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    Expand it. n!/n^n=(n/n)*((n-1)/n)*((n-2)/n)*...*(3/n)*(2/n)*(1/n). Does that give you any ideas?
     
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