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    A bounded positive multiplier preserves convergence of a positive series

    P-RA-WORKSHOP-D3-SEQ-12

    Area
    Real Analysis
    Topics
    Sequences of Numbers, Series of Numbers
    Status
    draft

    (January 2012 #1a) Let \(\{a_n\}\), \(\{b_n\}\) be bounded sequences of positive real numbers. If \(\sum b_n\) is convergent, show that \(\sum a_nb_n\) is also convergent.

    Solution

    1. Bound \(a_n\) uniformly.Proof: \(\{a_n\}\) is bounded and positive, so there is \(M > 0\) with \(0 < a_n \le M\) for all \(n\).

    2. Compare \(\sum a_n b_n\) with \(\sum b_n\).Proof: for every \(n\), \(a_n b_n \le M b_n\) (as \(b_n > 0\)). Since \(\sum b_n\) converges and \(M\) is constant, \(\sum M b_n\) converges; by the comparison test, \(\sum a_n b_n\) converges.

    3. Q.E.D.

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