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Values for b such that sum over b^log n converges

  1. Dec 11, 2008 #1
    What are all the positive values of "b" for which the series summation(b^log n) converges.

    I think we must find the values of b for which the sequence of partial sums form a bounded sequence. But it gets complicated when I try to solve the inequality
    "summation of k from 1 to n of (b^log k)< or equal to some M".
     
  2. jcsd
  3. Dec 11, 2008 #2

    HallsofIvy

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    Re: series

    Since those are all positive terms, I would be more inclined to use the ratio test: Look at b^log(n)/b^log(n+1) as n goes to infinity. You can use exponential and logarithm properties to simplify that.
     
  4. Dec 11, 2008 #3
    Re: series

    Thanks HallsofIvy, I tried that and I got 0<b<1 by just solving b^log(n)/b^log(n+1) <1. I am not sure how to introduce the limsup part. I mean, when I try to solve limsup [b^log(n)/b^log(n+1)] <1; I get lim as n-->inf of b^log(n+1/n) <1 which gives me b^ln1<1 and that gives me 1<1.
    How do I get rid of that contradiction?
     
    Last edited: Dec 11, 2008
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