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Help with convergence/divergence

  1. Oct 11, 2006 #1
    Sigma (-1)^n / (ln n)^n

    First I start using the ratio test (because I want to test if it's absolutely conver. or conditionally conver.), then I get the limit = 1 so it's inconclusive by the ratio test.
    But from the alternating series test, it's convergent. So it's conditionally convergent.

    Is my process right?
     
  2. jcsd
  3. Oct 11, 2006 #2
    limit absolute value of nln(n)/ (n+1)(ln(n+1))
    --->n/(n+1) so the limit is 1.
     
  4. Oct 11, 2006 #3
    yes you are correct. it is conditionally convergent, because it satisfies the alternating series test but fails the ratio test.
     
  5. Oct 11, 2006 #4

    StatusX

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    As you've written it in the first post, [itex](\ln(n))^n\neq n \ln(n)[/itex], it's [itex]\ln(n^n)= n \ln(n)[/itex]. Which did you mean?
     
    Last edited: Oct 11, 2006
  6. Oct 11, 2006 #5
    sigma n=1 to infinite 1/(n*ln(n))
    sigma n=1 to infinite (-1)^n /(n*ln(n))

    The second one is convergent, how about the first one? It's convergent as well?
     
  7. Oct 11, 2006 #6
    To StatusX,
    Find the limit for absolute value of nln(n)/ (n+1)(ln(n+1))
    I cross out ln(n) on the top with the ln(n+1) at the bottom, so it left with n/(n+1). Do you mean this one?
    For the first one, ln(n)^n is at bottom. So it's same as nln(n).
     
  8. Oct 11, 2006 #7
    [tex] \sum_{n=1}^{\infty} \frac{1}{n\ln n} [/tex].

    Use the integral test.


    [tex] \lim_{x\rightarrow \infty}\frac{n\ln n}{(n+1)(\ln(n+1))} = 1 [/tex]
     
    Last edited: Oct 11, 2006
  9. Oct 11, 2006 #8
    I see. It's divergent for the second one but is convergent for the first one because of Alternating series test.

    Thanks a lot.
     
  10. Oct 11, 2006 #9
    Here's another question.

    Sigma n=1 to infinity (-1)^(n-1)/n^p

    For what values of p is serie convergent?

    I have p>0, can p be 0 in this case?
     
  11. Oct 11, 2006 #10

    StatusX

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    I'm saying there's a difference between [itex](\ln(n))^n[/itex] and [itex]\ln(n^n)[/itex], and the one you've written in you're first post is not equal to [itex]n \ln(n) [/itex].
     
  12. Oct 11, 2006 #11
    [tex] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}} [/tex]. p cant be 0, because then [tex] a_{n} [/tex] wouldnt be decreasing, and it would fail the alternating series test. Thus [tex] p>0 [/tex].
     
  13. Oct 11, 2006 #12
    To StatusX,
    I see what you mean now.
    Let me re-do my problem.
    Thanks by the way.
     
  14. Oct 11, 2006 #13
    For [tex] \sum_{n=1}^{\infty}\frac{(-1)^{n} }{(\ln n)^{n}} [/tex], you could use the root test. So it is absolutely convergent.
     
  15. Oct 11, 2006 #14
    Thank you.
    Here's a question I don't understand.

    Suppose that sigma n=0 to the infinity Csubn*x^n converges when x=-4 and diverges when x=6. What can be said about the convergence or divergence of the following series?

    a) Sigma n=0 to infinity Csubn
    ..............

    Can someone give me a hint on this one?
     
  16. Oct 11, 2006 #15
  17. Oct 11, 2006 #16
    Thanks. But I'm not sure if I get the concept.
     
  18. Oct 11, 2006 #17
    a)con.
    b)div.
    c)con.
    d)div.
     
  19. Oct 11, 2006 #18
    yeah thats right
     
  20. Oct 11, 2006 #19
    I sort of get the concept...but not exactly sure. Thank you so much though.
     
  21. Oct 12, 2006 #20
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