Help with convergence/divergence

  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. limit absolute value of nln(n)/ (n+1)(ln(n+1))
    --->n/(n+1) so the limit is 1.
     
  4. yes you are correct. it is conditionally convergent, because it satisfies the alternating series test but fails the ratio test.
     
  5. StatusX

    StatusX 2,567
    Homework Helper

    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. 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. 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. [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. 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. 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. StatusX

    StatusX 2,567
    Homework Helper

    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. [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. To StatusX,
    I see what you mean now.
    Let me re-do my problem.
    Thanks by the way.
     
  14. 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. 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. Thanks. But I'm not sure if I get the concept.
     
  17. a)con.
    b)div.
    c)con.
    d)div.
     
  18. yeah thats right
     
  19. I sort of get the concept...but not exactly sure. Thank you so much though.
     
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