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Math Help - limits

  1. #1
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    limits

    PROBLEM STATEMENT:

    (a)Suppose that { a_n} and { b_n} are sequences of nonnegative numbers and let c_n = max(a_n,b_n). Show that \sum c_n converges iff both \sum a_n and \sum b_n converge. Hint: c_n \leq a_n+b_n.

    (b)Prove that if \sum a_n converges, where a_n \geq 0, then \sum \frac{\sqrt{a_n}}{n} converges, too. Hint: We already know that \sum \frac{1}{n^2} converges.

    MY WORK:

    (a) We use the comparison test here which says that if |a_n| \leq c_n for n \geq N_0, where N_0 is some fixed integer, and if \sum c_n converges, then \sum a_n converges.
    Supposing that \sum c_n converges, then the larger of a_n and b_n converges. Suppose a_n \geq b_n. Since \sum c_n converges, and c_n=a_n, \sum a_n converges. Now by the comparison test,  b_n converges. The same is true if b_n \geq a_n.
    Now, supposing that both \sum a_n and \sum b_n converge, since c_n \leq a_n and c_n \leq b_n, \sum c_n must converge by the comparison test.

    (b)Here, we use the Schwarz inequality which states that if a_1,...,a_n and b_1,...,b_n are complex numbers, then |\sum^n_{j=1} a_j\overline{b_j}|^2 \leq \sum^n_{j=1} |a_j|^2\sum^n_{j=1} |b_j|^2.
    From here, we directly get \sum \frac{\sqrt{a_n}}{n} \leq \sum a_n \sum \frac{1}{n^2} . and since both \sum a_n converges, and \sum \frac{1}{n} converges, then \sum \frac{\sqrt{a_n}}{n} converges.
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  2. #2
    Member
    Joined
    Sep 2010
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    86
    a) Rather prove it differently, your proof seems a little odd.

    You know that \sum a_n and \sum b_n converges, so \sum a_n + b_n converges.

    Since c_n = \max\{a_n,b_n\} you know that c_n \le a_n + b_n. Thus \sum c_n \le \sum a_n+ b_n and by the comparison test the first series must converge.

    b) Is perfectly fine.
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