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Does the limit converge or diverge

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    infinity
    Ʃ [itex]\frac{ln(n)}{n\sqrt{n}}[/itex]
    n=1

    2. Relevant equations
    I think that I need to use either the limit comparison test or the ratio test.


    3. The attempt at a solution

    After listing all the terms I found out that the function was positive and decreasing, and that the terms approached 0. However when I used the limit comparison test or the root test I keep on getting an answer of infinity or 0, which is inconclusive. So I don't know how else to approach the problem. Is there something easy that I am overlooking. Thanks.
     
    Last edited by a moderator: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2

    Mark44

    Staff: Mentor

    Show us what you tried for the limit comparison test. I agree that the ratio test would not be conclusive.
     
  4. Nov 6, 2011 #3
    I think the best test to use is the ratio test:

    If there exists a constant C < 1 such that |an+1/an|<C for all sufficiently large n, then ∑an converges absolutely.
     
  5. Nov 6, 2011 #4

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    Have you thought about the integral test?
     
  6. Nov 6, 2011 #5
    for the limit comparison test I tried a bunch of different comparisons. I said as lim n-> infinity of

    lim n→∞. of (ln(n)/(nsqrt(n))/(1/n^5/4).That didnt give me any help. Then I tried comparing it to 1/n^1/4. Then I tried comparing it to a bunch of other stuff but nothing seemed to work. I cant figure out what to do
     
  7. Nov 6, 2011 #6
    Also I never thought about the integral test. I will try that
     
  8. Nov 6, 2011 #7

    LCKurtz

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    @kuczmama: Don't use the X2 key inside tex brackets. Right click on this to see how to do exponents:

    [tex]n^{\frac 3 2}[/tex]
     
  9. Nov 6, 2011 #8
    I got that the integral converges to 4. I guess that this means the sum also converges. Thanks alot guys!!
     
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