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Proof of convergence

  1. Mar 15, 2005 #1
    How can I use the following definition to prove that the limit as n --> infinity of (n!)/(n^n) is 0 :

    the sequence {an}-->a if
    For all epsilon>0, there exists an N element of natural nos. such that for all n> or = to N, abs value(an-a)<epsilon
     
  2. jcsd
  3. Mar 15, 2005 #2

    HallsofIvy

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    That's really the hard way!

    Any way, notice that n!= n*(n-1)*(n-2)*...*(3)(2)(1) with n factors while
    nn= (n)*(n)*(n)*...*(n)(n)(n) also with n factors. That means that
    n!/nn can be written as a product of n factors: (n/n)((n-1)/n)((n-2)/n)...(3/n)(2/n)(1/n). Since the first term is 1 and all others are less than 1,
    n!/nn< 1/n. It should be easy to find N such that if n> N, |1/n|< epsilon!
     
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