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Convergence of Series

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the positive values of p for which the series converges.
    [tex]\Sigma_{n=2}^{\infty} \frac{1}{n( \ln (n)^{p})} [/tex]


    2. Relevant equations

    1/n^p converges if p>1, and diverges if =<1

    3. The attempt at a solution
    Don't know where to begin
     
  2. jcsd
  3. Mar 29, 2009 #2

    Dick

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    Try an integral test.
     
  4. Mar 29, 2009 #3
    Doesn't the integral tell me only whether it converges or diverges? I need to actually determine the values of p for which that will converge.
     
  5. Mar 29, 2009 #4

    Dick

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    Whether the integral converges or diverges will depend on the value of p.
     
  6. Mar 29, 2009 #5
    Is it two? Are we also assuming that p will be a positive integer rather than any positive real number?
     
  7. Mar 29, 2009 #6

    Dick

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    What do you get if you integrate 1/(x*(ln(x))^p)? Sure, take p to be positive. If it's not then the integral definitely diverges.
     
  8. Mar 29, 2009 #7
    http://texify.com/img/%5CLARGE%5C%21%5Cint_2%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7Bx%20%5Cln%28x%29%5E%7Bp%7D%7D%20%3D%20%20%20%5Cleft%5B%20%5Cfrac%7B%20%5Cln%20%28%5Cln%28x%29%5E%7Bp%2B1%7D%29%7D%7B%28p%2B1%29%7D%20%5Cright%5D_2%5E%7B%5Cinfty%7D.gif [Broken]
    What do I do now?
     
    Last edited by a moderator: May 4, 2017
  9. Mar 29, 2009 #8
    Have you considered working with a Taylor approximation of ln(1 + n).
     
  10. Mar 29, 2009 #9
    That is taught later in the chapter.
     
  11. Mar 29, 2009 #10
    What happens to ln(ln(x)^{p+1}) as x --> infty?

    OK, nevermind.
     
    Last edited by a moderator: May 4, 2017
  12. Mar 29, 2009 #11

    Dick

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    That's not right. Substitute u=ln(x).
     
    Last edited by a moderator: May 4, 2017
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