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Series question (non-baby rudin, ch. 4, #7)

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose [tex]\{a_n\}[/tex] is a sequence of positive numbers such that [tex]\sum_na_nb_n < \infty[/tex] whenever [tex]b_n \ge 0[/tex] and [tex]\sum_nb_n^2 < \infty[/tex]. Prove that [tex]\sum_na_n^2 < \infty[/tex].

    2. Relevant equations

    There's a suggestion in the text:
    If [tex]\sum_na_n^2 = \infty[/tex], then there are disjoint sets [tex]E_k \, (k=1,2,3,\dots)[/tex] so that
    [tex]\[\sum_{n\in E_k}a_n^2 > 1.\][/tex]
    Define [tex]b_n[/tex] so that [tex]b_n = c_ka_n[/tex] for [tex]n \in E_k[/tex]. For suitably chosen [tex]c_k[/tex], [tex]\sum_na_nb_n = \infty[/tex] although [tex]\sum_nb_n^2 < \infty.[/tex]



    3. The attempt at a solution
    Using the hint, we've got
    [tex]
    \[b_n^2 = c_k^2a_n^2; \qquad\qquad a_nb_n = c_ka_n^2\]
    \begin{align*}
    S1 = \sum_n b_n^2 &= \sum_{n\in E_1} c_1^2 a_n^2 + \sum_{n\in E_2} c_2^2 a_n^2 + \dots \\
    &= \sum_k \sum_{n_k\in E_k} c_k^2 a_{n_k}^2 < \infty\\
    S2 = \sum_n a_n b_n &= \sum_k \sum_{n_k\in E_k} c_k a_{n_k}^2 = \infty.
    \end{align*}
    [/tex]

    From this, it seems clear that we want to choose the [tex]c_k[/tex] to be "sufficiently dampening" so that [tex]S1[/tex] really is finite but [tex]S2[/tex] is not. I'm not entirely sure how to go about choosing the [tex]c_k[/tex] for arbitrary [tex]a_n[/tex] and [tex]b_n[/tex]. Thanks very much, any hints would be greatly appreciated! Sorry about the formatting, I'm new here..
     
  2. jcsd
  3. Nov 26, 2009 #2

    statdad

    User Avatar
    Homework Helper

    Perhaps a well-known inequality dealing with sums would help.
     
  4. Nov 26, 2009 #3
    Thanks for the hint! I'm assuming you're referring to the CBS inequality, but I'm not quite sure how to apply it in a useful way.

    I've got
    [tex]
    \sum_n a_nb_n \le \sqrt{\sum_n a_n^2}\sqrt{\sum_n b_n^2}
    [/tex]
    and the RHS is infinite since the sum of the squares of the a_n is infinite.
     
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