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Math Help - Multiplication of convergent series

  1. #1
    Senior Member I-Think's Avatar
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    Multiplication of convergent series

    Given that \sum{a_n} converges and [b_n] is monotonic and bounded, prove \sum{a_nb_n} converges.

    So I know a_n{\rightarrow{0}} and b_n converges, so a_nb_n{\rightarrow{0}}
    Where do I go from here?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Multiplication of convergent series

    Quote Originally Posted by I-Think View Post
    Given that \sum{a_n} converges and [b_n] is monotonic and bounded, prove \sum{a_nb_n} converges.

    So I know a_n{\rightarrow{0}} and b_n converges, so a_nb_n{\rightarrow{0}}
    Where do I go from here?
    PlanetMath: proof of Abel's test for convergence
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  3. #3
    Super Member girdav's Avatar
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    Re: Multiplication of convergent series

    Let S_n:=\sum_{k=1}^na_kb_k, s_k:=\sum_{j=1}^ka_k and M:=\sup_{j\in\mathbb N}|s_j|<+\infty since the sequence \{s_j\} is bounded. We have for n\geq 1:
    \begin{align*}S_n&=\sum_{k=1}^n(s_k-s_{k-1})b_k\\&=\sum_{j=1}^ns_jb_j-\sum_{j=0}^{n-1}s_jb_{j+1}\\&=\sum_{j=1}^ns_j(b_j-b_{j+1})+s_nb_n.\end{align*}
    We have to show that the sequence \{S_n\} is a Cauchy sequence. To see that, write for m,n\geq 1:
    \begin{align*}|S_{m+n}-S_n|&=\left|\sum_{j=m+1}^{m+n}s_j(b_j-_{j+1}) +s_{m+n}b_{m+n}-s_{m+1}b_{m+1}\right| \\&\leq\left|\sum_{j=m+1}^{m+n}s_j(b_j-_{j+1})\right|+\left|a_{m+n}b_{m+n}-a_{m+1}b_{m+1}\right|\\&\leq M|b_{m+1}-b_{m+n}|+\left|a_{m+n}b_{m+n}-a_{m+1}b_{m+1}\right|.\end{align*}
    Last edited by Plato; August 21st 2011 at 11:44 AM.
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