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Infinite series

  1. Mar 5, 2006 #1
    Find all [itex]p \geq 0[/itex] such that

    [tex]\sum_{k=1}^{\infty} \frac{1}{k \, (\log (k+1))^p}[/tex]

    converges.

    It looks like the integral test is the most likely candidate, but I haven't been able to make any progress using it. I'd appreciate a push in the right direction.

    Edit:
    I've managed to prove that it converges for [itex]p > 1[/itex]. Since it obviously diverges for [itex]p=0[/itex], I'm trying to see what happens when [itex]0 < p \leq 1[/itex].

    Edit2:
    And now I just proved that it diverges for such p. Problem solved. :smile:
     
    Last edited: Mar 5, 2006
  2. jcsd
  3. Mar 5, 2006 #2

    quasar987

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    out of curiosity, how did u proceed? I used a criterion that is very useful when dealing with log series. it says that [itex]\sum a_n[/itex] converges [itex]\Leftrightarrow \sum 2^n a_{2^n}[/itex] converges.

    So compare the [itex]2^n[/itex] serie with the riemann p-serie and you get that the original series behaves just like the riemann p-serie, i.e. diverges for [itex]p \leq 1[/itex] and converges for p>1.
     
    Last edited: Mar 5, 2006
  4. Mar 5, 2006 #3

    matt grime

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    How on earth does that test work?

    take the sum of (-1)^n/log(n), that converges by the alternating series test, yet the 2^n subseries

    2^n(-1)^(2^n)/log(2^n) = 2^n/nlog(2)

    does not converge.
     
  5. Mar 5, 2006 #4

    shmoe

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  6. Mar 5, 2006 #5

    quasar987

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    Oh yeah, an must be decreasing non-negative.
     
  7. Mar 5, 2006 #6

    shmoe

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    I was just about to add "non-increasing".

    For the problem here, you can also use the integral test.
     
  8. Mar 5, 2006 #7
    I used the fact that log is increasing so that [itex]\log (k) < \log (k+1)[/itex], and I used the integral test.
     
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