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Convergence here?

  1. May 20, 2006 #1
    The series is from k=2 to infinity.

    the term is 1/k(ln(k)).

    What I did was change the term to an integral... int(1/k(ln(k)) and solved it.

    I came up with ln(ln(k)).

    If I plug in infinty, then ln(ln(k)) goes to infinity so that means it diverges right.

    From the original graph of 1/k(ln(k)), it seems like it converges but it doesn't.

    Am I doing something wrong in my math? The original question asks if the series converges absolutely, conditionally, or diverges.

    The positive term (to check for absolute convergence) is already the whole term itself right?
     
  2. jcsd
  3. May 20, 2006 #2

    HallsofIvy

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    First an obvious point, 1/kln(k) is positive for all k>1 so if the series converges, it converges absolutely, not conditionally.

    [itex]\Sum \frac{1}{n}[/itex] diverges, [itex]\Sum \frac{1}{k^2}[/itex] converges and [itex]\frac{1}{k^2} \le \frac{1}{k ln(k)} \le \frac{1}{k}[/itex] so that (or looking at the graph) doesn't tell us anything!

    To integrate [itex]\int \frac{1}{x ln(x)}dx[/itex] let u= ln(x) (as I presume you did). Then [itex]du= \frac{1}{x}dx[/itex] so the integral becomes [itex]\int \frac{1}{u}du[/itex]= ln(u)= ln(ln(x)) as you say.
    That does not converge as x goes to infinity and so the series does not converge.
     
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