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Prove rigorously: lim (3^n)/(n!) = 0

  1. Jan 22, 2010 #1
    1. The problem statement, all variables and given/known data
    PROVE rigorously from the definition that
    lim (3n)/(n!) = 0.
    n->∞

    2. Relevant equations
    N/A

    3. The attempt at a solution
    By definition, a real number sequence
    a(n)->a iff
    for all ε>0, there exists an integer N such that n≥N => |a(n) - a|< ε.

    |(3n)/(n!)|<...< ε
    Now how can I find N? The usual approach to find N would be to set |a(n) -L|< ε and solve the inequality for n. But here in |(3n)/(n!)|<...< ε, I don't think we can solve for n. (because n is appearing everywhere. n(n-1)(n-2)...2x1, and n also appears on the numerator)

    Any help is greatly appreciated!



    [note: also under discussion in Math Links forum]
     
    Last edited: Jan 23, 2010
  2. jcsd
  3. Jan 22, 2010 #2

    Dick

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    Science Advisor
    Homework Helper

    Oh, come on. 3^6/6!=81/80. The next term multiplies that by 3/7. That factor is less than 1/2. The next term multiplies that by 3/8, that's also less than 1/2. The next term multiplies that by 3/9, also less than 1/2. Can you find an N such that (81/80)*(1/2)^N is less than epsilon? I think you probably can. Call off the debate.
     
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