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Convergence of a particular infinite sum

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex]b_n[/itex] be a bounded sequence of nonnegative numbers. Let r be a number such that [itex]0 \leq r < 1[/itex].
    Define [itex]s_n = b_1*r + b_2*r^2 + ... + b_n*r^n[/itex], for all natural numbers n.
    Prove that [itex]{s_n}[/itex] converges.


    2. Relevant equations
    Sum of first n terms of geometric series = [itex]sum_n = (a_1)(1-r^{n+1})/(1-r)[/itex]


    3. The attempt at a solution
    Clearly, [itex]{s_n}[/itex] is monotonically increasing.
    Since [itex]{b_n}[/itex] is bounded, [itex]|b_n| \leq M[/itex], for all natural numbers n.

    I want to use the fact that if [itex]{s_n}[/itex] is both monotonically increasing and is bounded, then it must converge. The part of the problem that has stumped me for the past 45 minutes is how to show that [itex]{s_n}[/itex] is bounded.

    The only material we've covered regarding infinite series thus far is for purely geometric series, which doesn't fit this problem precisely--but I included the formula anyway.

    Help would be greatly appreciated! I have an exam on Monday and getting so completely stymied by a simple problem is not doing wonders for my confidence.
     
  2. jcsd
  3. Sep 10, 2011 #2

    Dick

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    sn=b1*r+b2*r^2+...bn*r^n<=M*r+M*r^2+...+M*r^n, right? That's just a hint. Does it help?
     
  4. Sep 10, 2011 #3
    Aghhhh, so simple! Something about the way I do proofs is just <i>wrong</i>, I always get caught in roadblocks of thinking and miss simple detours like that.

    [itex]s_n \leq M*r + M*r^2 + ... + M*r^n[/itex]
    Then [itex]s_n \leq M(r-r^{n+1})/(1-r)[/itex]
    Then [itex]s_n \leq Mr/(1-r) = M'[/itex]
    Then [itex]|s_n| \leq M'[/itex] for all natural numbers n, so [itex]s_n[/itex] is bounded.
    Thus, since [itex]s_n[/itex] is bounded and monotonically increasing, we have [itex]s_n[/itex] converges.

    Or, I think that's right, anyway.

    Thanks a bunch!
     
  5. Sep 10, 2011 #4

    Dick

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    It's exactly right. Good take on the hint.
     
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