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Prove series converges

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data

    If a_1 + a_2 + ... is an infinite series converging to A, and b1, b2, ... is an infinite sequence that is bounded and monotonic, prove that (a_1)(b_1) + (a_2)(b_2) + ... converges


    2. Relevant equations



    3. The attempt at a solution

    I don't really know where to start...all I can say is that if a_1 + a_2 + ... converges, then a_n approaches 0 as n goes to infinity, and so (a_n)(b_n) also has a limit of 0, since b_n converges to some finite value.
     
  2. jcsd
  3. Jun 22, 2009 #2

    Office_Shredder

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    You know that bn is bounded and monotonic. Have you tried using that? Given a general number b, you know
    [tex] \sum a_n b [/tex]
    converges right? Now try to compare that to the sequence

    [tex] \sum a_n b_n[/tex] choosing b such that |bn| < b for all n
     
  4. Jun 22, 2009 #3

    tiny-tim

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    Hi JG89! :smile:
    Yes :approve: … concentrate on that value (call it b) …

    then use deltas and epsilons. :wink:
     
  5. Jun 23, 2009 #4
    I have an intuitive idea of what's going on, but I'm having a hard time fleshing it out into an epsilon argument.

    First off, since a_1 + a_2 + ... converges then b(a_1 + a_2 + ...) converges. I know that for large enough n, |(a_n)(b_n)| gets 'really' close to |(a_n)b|, since b is the limit of b_n. Since the b_n are monotonic, then as n increases the |(a_n)(b_n)| gets even closer to |(a_n)b|. I can in fact make this difference as small as I please, provided n is taken large enough and so it seems that after a certain n, the difference between (a_n)(b_n) and (a_n)b will become "negligible" and since (a_n)b is Cauchy, then (a_n)(b_n) is also Cauchy.
     
    Last edited: Jun 23, 2009
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