1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    (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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook