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Homework Help: Root Test, Infinite Series

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

    I don't have the problem in front of me but it was something like "converge or diverge"?:

    [tex]\sum[/tex] 5^n/(n+1)





    3. The attempt at a solution

    I would like to know that if I use the root test, would I get lim n-> [tex]\infty[/tex] 5/(n+1)^1/n = 5/(n+1)^0 = 5/1 = 5 ?

    I suspect this is not correct since (n+1)^1^n is an indeterminate form, however, in the book the only way to solve the problem that led to the correct answer was if (n+1) became 1

    (this is not the exact example, I am more interested in if you can use the root test on variables variables of n in the base, ex lim -> [tex]\infty[/tex]n^1/n




    If this is not clear enough I will re-post with the correct problem.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 1, 2010 #2

    Dick

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    (n+1)^(1/n) may be indeterminant, but the limit as n->infinity of (n+1)^(1/n) is 1. Take the log and apply l'Hopital.
     
  4. May 1, 2010 #3
    do you mean, y=(n+1)^1/n
    lny = ln(n+1)^1/n,
    lny = (n+1)/n

    y'/y = 1/1
    e^y'=e^y... ok I got kinda lost
     
  5. May 1, 2010 #4

    Dick

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    ln((n+1)^(1/n))=(1/n)*ln(n+1)=ln(n+1)/n. Apply l'Hopital. What's limit ln(n+1)/n?
     
  6. May 1, 2010 #5
    ok, but when I take the derivative of the top and bottom with respect to n I get (1/(n+1)) / 1 = 1/(n+1), and that -> 0 as n - > infinity. Not sure how you are getting 1 as an answer.
     
  7. May 1, 2010 #6

    Dick

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    That limit is telling you that log(y)=0. Remember, you took the log? What's y?
     
  8. May 1, 2010 #7
    so take e on both sides and you get y = e^0, or y = 1

    :)


    Thanks! I think that's just something worth memorizing!
     
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