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Prove that a sequence is eventually decreasing.

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that if a > 0, the sequence [tex]\frac{a^{n}}{n!}[/tex] is eventually decreasing for large n.


    2. Relevant equations
    N/A


    3. The attempt at a solution

    I know I'm to prove this by finding the index but I'm at a loss on how to find the index on such a sequence as I'm not used to using a variable that's in the sequence.

    To warmup I was given a values 110 and 300 to find an index for but with 'a' being in the sequence itself I'm not sure how to do that.
     
  2. jcsd
  3. Feb 12, 2010 #2

    Gib Z

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    Homework Helper

    Try showing that [itex]n! - a^n[/itex] is eventually increasing.
     
  4. Feb 13, 2010 #3

    HallsofIvy

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    The numerator and denominator are products of n terms but each term in the numerator is a while the terms in the denominator are increasing. The "average" value (geometric average since we are multiplying) in the denominator is [itex]\sqrt[n]{n!}[/itex]. Can you show that, for fixed a, there exist an integer, n, such that [itex]\sqrt[n]{n!}> a[/itex]?
     
  5. Feb 14, 2010 #4
    I don't really follow.
     
  6. Feb 14, 2010 #5
    try subtracting the n+1 term of the sequence from the nth term. if this quantitiy is less that zero the the sequence is decreasing. however you can factor out a to the power n. this is greater than zero so it doesn't matter. look at the other factor, if you choose a <N+1 you get the desired result.
     
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