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Is this proof for convergence of 3^n/n! rigorous enough?

  1. Sep 16, 2012 #1
    Hi,

    I am trying to self study analysis and was practicing some problems. I wasn't sure if this solution to one of the problems I came across was rigorous enough.

    Basically, by writing down the first few terms of 3^n and n!, I figured I can say 3^n < 3*(n-1)! for all n>=13...without actually calculating the numerical values.

    so (3^n/n!)<3/n for n>=13. Therefore, for any ε we can can find N>13 and N>3/ε. And therefore the term 3^n/n! converges to 0. Is this a right solution? Is it rigorous enough? Could you advise me of a better way to prove this?

    Thanks.
     
    Last edited: Sep 16, 2012
  2. jcsd
  3. Sep 16, 2012 #2

    Ray Vickson

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    You can prove it by induction. First, you can establish it is true for n = 8. So assuming it is true for n = k >= 8, prove it is also true for n = k+1. (This is EASY.) Thus, it is true for all n ≥ 8.

    RGV
     
  4. Sep 16, 2012 #3
    I assume, you mean, to prove that 3^n<3(n-1)! for all n>=8. I guess that makes sense..Thanks.
     
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