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Thread: limits

  1. #1
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    limits

    PROBLEM STATEMENT:

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

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

    MY WORK:

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

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

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

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

    b) Is perfectly fine.
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