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Analysis Proof (simple, but !)

  1. Jun 30, 2008 #1
    Analysis Proof (simple, but urgent!)

    1. The problem statement, all variables and given/known data

    This is an exercise from Rudin Chapter 3.

    #8. If [tex]\sum a_{n}[/tex] converges, and if {bn} is monotonic and bounded, prove that [tex]\sum b_{n} a_{n}[/tex] converges.

    2. Relevant equations

    THEOREM: Suppose {sn} is monotonic. Then {sn} converges <==> {sn} is bounded.

    3. The attempt at a solution

    {bn} converges. bn --> b
    So, for any [tex]\epsilon > 0 \exists N [/tex] such that n > N implies [tex]|b - b_{n}| < \epsilon[/tex]

    Let M = b if b > bN
    Let M = [tex]\epsilon + b[/tex] if b <= bN

    Now, [tex]\sum M a_{n} = M \sum a_{n}[/tex] for M constant. And [tex]\sum M a_{n} [/tex] converges to Ma. So, [tex]\sum b_{n} a_{n}[/tex] converges.


    -----

    Am I on the right track with this? I need to turn this in soon and any help would be GREAT!
     
  2. jcsd
  3. Jun 30, 2008 #2

    dynamicsolo

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    Homework Helper

    Re: Analysis Proof (simple, but urgent!)

    Maybe one of the mathematicians here will have more to say about this, but it strikes me that there are two cases to consider, since the proposition only says {b_n} is monotonic and bounded.

    What you've treated is the case where {b_n} is monotonically increasing, so since b_n never gets bigger than a constant M, and the sum of M·a_n will converge, so will the sum of b_n · a_n .

    You may also need to look at the case where {b_n} is monotonically decreasing. The basic argument is similar, but one other remark will need to be included...
     
  4. Jun 30, 2008 #3
    I didn't fully understand what you were doing, but it looks like that you are trying to use the knowledge that the sequence [tex]b_n[/tex] is converging. You should not try to prove this claim:

    "Suppose that the series [tex]\sum_n a_n[/tex] converges, and that the sequence [tex]b_n[/tex] converges. Then the series [tex]\sum_n b_n a_n[/tex] converges."

    That is not right: Set

    [tex]
    a_n = \frac{(-1)^n}{\sqrt{n}}
    [/tex]

    [tex]
    b_{2n+1} = 0,\quad b_{2n}=\frac{1}{\sqrt{n}}.
    [/tex]

    for a counter example.

    Sorry, I don't know how to do that exercise, and cannot give more advice now, but I know that you must use the knowledge that [tex]b_n[/tex] is monotonic and bounded for something more than the convergence of [tex]b_n[/tex].
     
  5. Jun 30, 2008 #4
    Re: Analysis Proof (simple, but urgent!)

    Thank you! That was helpful, I think I have a better proof now!
     
  6. Jun 30, 2008 #5

    dynamicsolo

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    Re: Analysis Proof (simple, but urgent!)

    Isn't [tex]b_{n}[/tex] supposed to be monotonic for this proposition?
     
  7. Jun 30, 2008 #6
    Re: Analysis Proof (simple, but urgent!)

    That was a counter example, I think. Showing that bn being monotonic matters.
     
  8. Jun 30, 2008 #7
    Re: Analysis Proof (simple, but urgent!)

    Consider two cases where b is a monotonely increasing sequence and monotonely decreasing.

    Then you can use the Theorem 3.42 in Rudin's book to show that series [tex]\sum_n a_n \left(b_{n}-b\right)[/tex] converges, then [tex]\sum_n a_n b_n [/tex] converges, where b is the limit of the sequence [tex] b_{n} [/tex]

    Hint:[tex]\lim_{n \to \infty} \left(b_{n}-b\right) =0 [/tex]
     
    Last edited: Jun 30, 2008
  9. Jun 30, 2008 #8

    dynamicsolo

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    Re: Analysis Proof (simple, but urgent!)

    Right, sorry... On re-reading the entire thread, the context of some of the posts is clearer to me. I see the theorem jostpuur was remarking on. I was focused on the boundedness and thought that was what you were using in your argument. The theorem is not required for the proof.
     
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