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Converge or diverge

  1. Feb 21, 2008 #1
    I know it is easy but I need help. lim(n->infinity)((-1)^n)/n diverge or converge?and why?

    p.s: question may not be understandable. Sorry for that.
  2. jcsd
  3. Feb 21, 2008 #2
    what are your thoughts?? Show what you did so far!
  4. Feb 21, 2008 #3
    there's a test for this type of series
  5. Feb 21, 2008 #4
    Well, in fact I did not do so much. I could not decide whether it converges 1 or -1 or since we do not know where it converges if we can simply say that it is diverge.
  6. Feb 21, 2008 #5
    why in the world would you think it converges to 1 or -1
  7. Feb 21, 2008 #6
    I was talking about (-1)^n, when I said it converges 1 or -1.
    Last edited: Feb 21, 2008
  8. Feb 21, 2008 #7

    Couldn't you just evaluate the series this way?
  9. Feb 21, 2008 #8
    And get what?????
  10. Feb 21, 2008 #9
    Are you just looking whether that limit exists, or your original question requires sth else, like whether the series with the general term a_n=(-1)^n /n converges or not?

    WHat is your original question.????
  11. Feb 21, 2008 #10
    This is the original question. If you want to check it, you can look at p.757 of Thomas's calculus (question 24). It says that find whether this sequence is convergent or not. If convergent find the limit of it?
  12. Feb 21, 2008 #11
    Well, to show that a series is convergent you first need to determine whether it is monotonic or not, then show that it is bounded. Or if the limit of that sequence exists then from a theorem we conclude that it is convergent also.
    To evaluate that limit you might want to breake the problem into two parts.
    First let n=2k, where k is any integer, and see where does that sequence tend to as k-->infinity.
    Second take n=2k+1, or n=2k-1, where again k is any integer, and see where does that sequence tend to, as k-->infinity. If they match then the lim exists, so the sequence converges to that value.
    there are other ways to show this though. Others might present you to those ways!
  13. Feb 21, 2008 #12
    Or I assume you could use a test for convergence
  14. Feb 21, 2008 #13
    Are you asking if this converges:

    [tex]\lim_{n \to \infty} \frac{(-1)^n}{n}[/tex]

    Look at the first few terms:

    [tex]\frac{-1}{1}, \frac{1}{2}, \frac{-1}{3}, \frac{1}{4}, \ldots[/tex]

    Do those numbers tend to get close to something?

    Other people here seem to ask if you're wondering if this converges:

    [tex]\sum_{n=1}^\infty \frac{(-1)^n}{n}[/tex]

    Have you looked at the first few terms of that?
  15. Feb 21, 2008 #14
    An alternating series will converge if it deceases at least as quickly as 1 over n. (see alternating series test)
  16. Feb 22, 2008 #15
    1/n remains always positive but tends to 0 as n tends to infinity.
    Using Leibnitz's test, convergence of the series you mentioned follows.

    Or you could apply the Dirichlet's test.
    The sequence of partial sums of (-1)^n is bounded, and 1/n remains always positive but tends to 0 as n tends to infinity.
    So convergence follows naturally.
    Last edited: Feb 22, 2008
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