1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Absolute convergence

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\sum_{n=2}^{\infty} \ln \left(1+\frac{(-1)^n}{n^p}\right)[/tex]

    p is a real parameter, determine when the series converges absolutely/non-absolutely

    3. The attempt at a solution

    I tried to do the limit [tex]\lim_{n\rightarrow \infty} \frac{\ln \left(1+\frac{(-1)^n}{n^p}\right)}{\frac{(-1)^n}{n^p}}[/tex], which is equal to one and this suggests that the series coverges if p is positive (limit comparison test). But then I'm not sure how to determine the absolute/non-absolute convergence. Could you help me please? Thanks very much in advance!
    Last edited: May 6, 2008
  2. jcsd
  3. May 6, 2008 #2


    User Avatar
    Gold Member

    Try to split the series into two parts - the even numbers and the odd ones and see if they both converge.
  4. May 7, 2008 #3
    i made a mistake in the first post, the limit comparison test is applicable only to non-negative series. then the limit should be like:
    [tex]\lim_{n\rightarrow \infty} \frac{\left|\ln \left(1+\frac{(-1)^n}{n^p}\right)\right|}{\frac{1}{n^p}}[/tex]

    which is equal to zero when p is positive. However, the lower series converges when p>1. Therefore, the original logarithm series converges absolutely for p>1.

    The textbook then says that the series converges non-absolutely also for 1/2 < p <= 1. But I can't prove it - do you have any ideas?
    Last edited: May 7, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook