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Homework Help: Non-absolutely convergent proof, help please

  1. Nov 29, 2006 #1
    The problem states:
    Suppose [itex]\sum a_n[/itex] and [itex]\sum b_n[/itex] are non-absolutely convergent. Show that it does not follow that the series [itex]\sum a_n b_n[/itex] is convergent.

    I tried supposing that the series [itex]\sum a_n b_n[/itex] does converge, to find some contradiction. So the series satisfies the cauchy criterion and the definition of convergence. I can't break the series apart (or can I?) so this is where I get stuck.

    Then I wrote the implications of the first sentence to try to come up with a statement that doesn't allow [itex]\sum a_n b_n[/itex] to be convergent. I get stuck again.

    What does a series being non-absolutely convergent imply that is useful?

    Is it true that [itex]\sum |a_n b_n|[/itex] < [itex]\sum |a_n|[/itex] [itex]\sum |b_n|[/itex] ? I don't know if that would help

    Sorry it looks like I don't have much work done, but I've been looking at this for several days.

    Note: The section in which the problem is assigned talks about the boundedness criterion for convergence, the Cauchy criterion for convergence, and absolute convergence, so I was hoping to come up with a proof that uses the information from the section.

  2. jcsd
  3. Nov 30, 2006 #2


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    Why would you do that? Do you know what you're being asked to prove?
    Yes, it's obviously true (although it should be <, not <), and no, obviously it doesn't help.
    You need to know what you're trying to prove first. Basically, you want to find an example of series [itex]\sum a_n[/itex] and [itex]\sum b_n[/itex] such that:

    a) both converge
    b) neither converge absolutely
    c) [itex]\sum a_nb_n[/itex] doesn't converge

    What types of series converge but don't converge absolutely? Ones that have some positive terms and some negative terms. Hint: take [itex]a_n = b_n[/itex]. Then once you prove [itex]\sum a_n[/itex] converges non-absolutely, you've automatically proven that [itex]\sum b_n[/itex] converges non-absolutely. Moreover, if you do this, then you get:

    [tex]\sum a_nb_n = \sum a_n^2[/tex]

    a sum of positive numbers. So whereas [itex]\sum a_n[/itex] is supposed to be a series that converges, but doesn't converge absolutely, hence converges only because it has negative terms "balancing out" its positive terms, the series [itex]\sum a_n^2[/itex] has no negative terms, so it's "more likely" to be divergent. What's a very common example of a divergent series?
    You don't need any of that information.
  4. Nov 30, 2006 #3


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    AKG's point is that assuming [itex]\sum a_n b_n[/itex] does converge would be perfectly reasonable if you were trying to prove that, under the given hypotheses, [itex]\sum a_n b_n[/itex] never converged. But that is not the case. You want to show that the statement "If [itex]\sum a_n[/itex] and [itex]\sum b_n[/itex] converge then [itex]\sum a_n b_n[/itex]" is NOT true. You want to find a counter example.
  5. Nov 30, 2006 #4
    Have you both never heard of a proof by contradiction??

    A statement can't be proven true by example.
  6. Nov 30, 2006 #5


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    See here.
    additional characters
  7. Nov 30, 2006 #6


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    Did you not READ what we both said? The original problem was "Suppose [itex]\Sum a_n[/itex] and [itex]\Sum bn[/itex] are non-absolutely convergent. Show that it does not follow that the series [itex]\Sum a_nb_n[/itex] is convergent." It does NOT follow. In other words prove that it is not true. You certainly can use a counter-example to prove something is NOT true.
  8. Nov 30, 2006 #7


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    Arcadiaz, prove that a natural number N is not necessarily even.

    If you cite 3 as an example that isn't even, you've shown that N is not necessarily even.
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