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Limit with factorial

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Why does the limit as n -> infinity of [3^(n+1)]/(n+1)!] * n!/(3^n) equal
    the limit as n -> infinity of 3/(n+1)?

    2. Relevant equations



    3. The attempt at a solution
    I have never encountered this before.
     
  2. jcsd
  3. May 8, 2007 #2

    Curious3141

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    Homework Helper

    Forget about the limit and just focus on simplifying algebraically. You know that [tex]{(n+1)}! = n!{(n+1)}[/tex]. Also [tex]3^{n+1} = (3)(3^n)[/tex]. Use those to cancel some terms and see what you get.
     
  4. May 8, 2007 #3
    Forget about the limit, how do you simplify:

    [3^(n+1)]/(n+1)! * [n!/(3^n)]

    (LOL, Curious3141 is faster than me)
     
  5. May 8, 2007 #4

    Curious3141

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    And it's weird how we worded that almost identically! :eek:
     
  6. May 8, 2007 #5
    Ah yes, that makes perfect sense. Thanks.
     
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