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Ratio Test Problem

  1. Nov 7, 2012 #1

    Ʃ (((n)!)^3)/(3(n))! Use the ratio test to solve
    n=1


    So first i put it into form of (n!)^3/3n!, then applied ratio test.

    from ratio got ((n+1)!)^3/(3n+1)! times (3n)!/(n!)^3

    from here im on shaky ground

    i go reduce the terms to (n!)^3(n+1)^3/(3n!)(3n+1) times (3n!)/(n!)^3

    i then end with lim n->∞ (n+1)^3/(3n+1)

    My main two questions are did i reduce terms correctly and cancel, and how to take the limit of the remaining term.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 7, 2012 #2
    Forgot to mention on my calculator it says the limit goes to infinity but i need to know how to actually show steps, etc
     
  4. Nov 7, 2012 #3

    Dick

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    You are being a bit too sloppy. There is a difference between 3n+1 and 3(n+1).
     
  5. Nov 7, 2012 #4
    Sorry i attached a pic of the actual prob if that would help

    Also i dont see where i have that problem i took (3n+1)! to (3n!)(3n+1)
     

    Attached Files:

  6. Nov 7, 2012 #5

    Dick

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    You do have that problem. (3n+1)!/(3n)! is (3n+1). (3(n+1))!/(3n)! is not the same thing.
     
  7. Nov 7, 2012 #6
    Okay i worked it again and tried to follow what you said hopefully

    I attached a pic of my work, srry for the sloppiness :P
     

    Attached Files:

  8. Nov 7, 2012 #7

    Dick

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    Yeah, I think you've got it. The limit isn't infinity, right?
     
  9. Nov 7, 2012 #8
    No im getting 1/27 when i divided the highest terms
     
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