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First test of divergence lim n-> n / 8^n

  1. Apr 30, 2007 #1
    1. The problem statement, all variables and given/known data
    n / 8^n

    2. Relevant equations

    3. The attempt at a solution
    It converges to 8 / 49? Not sure how.

    First test of divergence
    lim n-> n / 8^n. infinity / infinity = 1. BUT bottom grows fast.
    Using L`Hospital
    lim n -> 1 / 3*8^n*ln(2) ---> goes to 0

    Tried to use the ratio test

    [ 1 / 3*8^(n+1)*ln(2) ] * [ 3*8^n*ln(2) ]
    1/8 is left from the ratio test
  2. jcsd
  3. Apr 30, 2007 #2
    What do you mean by "n / 8^n".
    This is not a clear problem statement.
    And how did you come to guessing the solution as 8/49?

    Do you mean you have to calculate Sum[n / 8^n,{n,0,Infinity}] ?
    If this is what you mean, then indeed 8/49 is the correct answer.
    (I did more effort to guess what you meant than yourself to explain your need)

    My hint:

    Sum[1 / 8^n,{n,0,Infinity}] is an easy series
    Sum[1 / 8^(x*n),{n,0,Infinity}] is just as easy
    Sum[n / 8^(x*n),{n,0,Infinity}] is related to the previous one by a simple operation
    Sum[n / 8^n,{n,0,Infinity}] is a special case of the previous
    Last edited: Apr 30, 2007
  4. Apr 30, 2007 #3


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    That's cute. I would have used (x/8)^n, myself.
  5. Apr 30, 2007 #4
    this series obviously converges.
    Here is a little analysis on how to come to this conclusion.
    Are you familiar with Dallambers covergence test( i am not sure if it is spelled dallamber, however)?
    It states that if limn->infinity a_(n+1)/a_n is larger than 1 it diverges, if it is smaller than 1, and positive it converges. so

    lim n->infinitya_n+1/a_n=lim n->infinity[8^n(n+1)]/[8n*8^n]=lim n->infinity(n+1)/8n=1/8<1, so the series converges
  6. Apr 30, 2007 #5
    d'Alembert's ratio test
    see wiki
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