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Radius & interval of convergence of power series

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    i was doing this exercise and came across this example.

    Ʃ (x^n)/ln(n+1)
    n=1



    3. The attempt at a solution
    i know you have to do the ratio test which is
    lim | a(n+1)/a(n)|
    n>∞

    i got to

    lim | [x ln(n+1)] /ln(n+2) |
    n>∞

    and have no idea how to continue? is there a way to get rid of the logs? since in this case as n>∞ the top and bottom just goes to ∞?
     
    Last edited: Oct 30, 2011
  2. jcsd
  3. Oct 30, 2011 #2
    ln(n)/ln(n+1) goes to 1 for n -> infty
    You can prove that easily by proving that ln(n)/ln(n+1)-1 approaches zero.
     
  4. Oct 30, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Or use L'Hopital's rule: the derivative of ln(x+1) is 1/(x+1) and the derivative of ln(x) is 1/x so this becomes (1/x)(x+1/1)= (x+1)/x goes to 1 as x goes to infinity.

    That shows that ln(x)/ln(x+1) goes to 1 as x goes to infinity which includes n going to infinity.
     
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