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Homework Help: Infinite series sum of (-1)^n/lnx

  1. Jul 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Find if the series is absolutely convergent, conditionally convergent, or divergent.

    The sum from two to infinity of (-1)^n/lnx.

    2. Relevant equations

    3. The attempt at a solution
    I don't know how to integrate 1/lnx, so that failed. The ratio and root test don't seem to simplify matters any further. I wasn't sure of anyway I could use the comparison test or limit ratio test. I used Alternate Series Test so I at least know that it's convergent, but I don't know if it's conditionally convergent or absolutely convergent.
  2. jcsd
  3. Jul 21, 2008 #2
    Try thinking some more about comparision tests.
  4. Jul 21, 2008 #3


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    That is an alternating series (that's crucial) and ln(x) is an increasing function. What does that tell you?
  5. Jul 21, 2008 #4
    Hm, would it be reasonable to say that because n > ln(n) for all n from 2 to infinity, 1/ln(n) > 1/x, which is divergent, thus making 1/ln(n) divergent and the sum from 2 to infinity of ((-1)^n)/ln(n) conditionally convergent? It didn't really occur to me to compare a transcendental function to an algebraic function (if it's even legal, anyway). Thanks for the help (assuming my solution was okay)!
    Last edited: Jul 21, 2008
  6. Jul 21, 2008 #5


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    Yes, the "alternating series" test says simply that if terms of [itex]\sum a_n[/itex] are alternating in sign and |an| is decreasing, then the series converges. In fact, it is easy to see that, given any N, the entire sum lies between the partial sum up to N and the partial sum up to N+1.
  7. Jul 21, 2008 #6
    But would that test be useful if you're distinguishing between whether something is absolutely convergent or conditionally convergent?
  8. Jul 21, 2008 #7


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    It will possibly tell you if it's conditionally convergent. You'll need another test to answer whether is absolutely convergent or not. Something like the integral test or a comparison test.
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