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Absolute convergence

  1. Mar 5, 2008 #1
    If [tex]\sum x_n[/tex] converges absolutely, and the sequence (yn) is bounded, then the sum [tex]\sum x_n y_n[/tex] converges.

    Find a counterexample that shows this isn't true when [tex]\sum x_n[/tex] is conditionally convergent.

    I'm honestly not to sure where to begin with this one. I was thinking Monotone Convergence Theorem, but that might not be necessarily true for xn Any suggestions would be fantastic!
  2. jcsd
  3. Mar 5, 2008 #2
    Here's one way. Take some concrete example of alternating series x_n, that converges, but not absolutely. Then start choosing values for y_n one by one, so that |y_n| < 1, and trying to make the value of the series x_n*y_n as large as possible. You should be able to get infinity out of it.
  4. Mar 5, 2008 #3
    I'm not entirely sure I follow you...
    Are you suggesting I choose lim(sup{xn}) or lim(inf{xn}, which are both monotone decreasing and increasing respectively?
  5. Mar 5, 2008 #4
    Wow, ok nevermind:

    We know [tex]\sum x_n[/tex] converges.
    |yn| <= M, For any n contained in the naturals - Def of bounded.

    [tex]\sum x_n y_n = \sum \left|x_n\right| \left|y_n\right| \leq M\sum \left|x_n\right|[/tex]

    Therefore, the sum converges!

    A counter example would be to let xn = [tex]\frac{(-1)^n}{n}[/tex] and yn = [tex](-1)^n[/tex]
  6. Mar 5, 2008 #5
    There is a really easy way to do this problem. What is one series you know that diverges, the one I am thinking of is probably one of the first examples you received as a divergent series when you first learned about series. Now is there a series that looks very similar to this one that converges, but not absolutely (perhaps an alternating series...). Then how can you turn this alternating series into the one which does not converge by choosing appropriate, and bounded yn?

    Edit: Nevermind you figured it out.
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