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Rudin convergence proof

  1. Jul 21, 2010 #1
    1. The problem statement, all variables and given/known data

    If [tex]\Sigma[/tex]an is a convergent series and {bn} is a monotonic and bounded sequence, then [tex]\Sigma[/tex] anbnis a convergent series.

    2. Relevant equations

    3. The attempt at a solution

    Since {bn} is bounded, |bn|<M for some M>0. Since [tex]\Sigma[/tex]an is a convergent series, we have that for every [tex]\epsilon[/tex]>0, there is some N>0 such that |Am-An|<[tex]\epsilon[/tex]/M for all m>n>N. Thus, [tex]\sum_{k=n}^m[/tex] akbk < M [tex]\sum^{m}_{k=n}[/tex] (Ak-Ak-1) < [tex]\epsilon[/tex].
    \Sigma [/tex] ak bk
    is convergent.

    Is this correct? If it is, then where did the monotonic behavior of {bn} get put in the proof?
  2. jcsd
  3. Jul 21, 2010 #2


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    Your proof doesn't make any particular sense. But I think the type of case they are thinking of is an=(-1)^n/n and bn=(-1)^n. The series an is conditionally convergent, but an*bn is divergent.
  4. Jul 21, 2010 #3
    I used the Cauchy criterion for convergence. Where does it not make sense?

    {bn} is a monotone sequence by hypothesis so bn = (-1)^n wouldn't work for this.
  5. Jul 21, 2010 #4


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    It's works better than I thought it did now that I realize An is the nth partial sum of an. You should define your terms. But it still only works for an absolutely convergent series. I KNOW (-1)^n isn't monotone. You were asking why you need to assume bn to be monotone. I'm giving you a non-monotone example where your proof fails and asking you to try to figure out why it fails. Why do you need the monotone assumption?
  6. Jul 21, 2010 #5
    I didn't think my proof was correct (since I don't see the use of the monotone assumption) so I was asking what was wrong with my proof; but if it was somehow correct, where is the monotone behavior hidden. I already know a proof for this but I was wondering why this particular one didn't work.
  7. Jul 21, 2010 #6


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    It's because of signs. (-1)<1. (-2)<(-1). But (-2)*(-1) is not less than 1*(-1). Do you see it now? You are assuming everything is positive.
  8. Jul 21, 2010 #7
    Ahh. I see. Thanks. I've been obsessing over this for quite some time.
  9. Jul 22, 2010 #8
    You are going to have to use Abel's summation by parts.

    This is a pretty hard question. I remember being stuck on it last year.
  10. Jul 22, 2010 #9


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    What about using the sandwich theorem?
  11. Jul 22, 2010 #10
    I think you can use theorem 3.42 if you can find a way to define a new sequence c_n from b_n. I have that written in some old notes and if I didn't make a mistake the proof isn't long.
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