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Relation of two complex series

  1. Aug 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that ##\left\{a_n\right\}## is a sequence of complex numbers with the property that ##\sum{a_n b_n}## converges for every complex sequence ##\left\{b_n\right\}## such that ##\sum{|b_n|^2}<\infty##. Show that ##\sum{|a_n|^2}<\infty##.


    2. Relevant equations



    3. The attempt at a solution

    We know that ##\lim{a_n}=0##, since if that were not the case then for ##b_n=\frac{1}{n a_n}##, ##\sum{|b_n|^2}<\infty##, but ##\sum{a_n b_n}=\sum{\frac{1}{n}}## diverges. Not sure that's helpful though.


    My other thought was to try to prove the contrapositive, that given ##\left\{a_n\right\}## such that ##\sum{|a_n|^2}## diverges, we could find a ##\left\{b_n\right\}## such that ##\sum{a_n b_n}## diverges as well.

    So we can find a sequence ##\left\{n_k\right\}## such that ##\sum_{n_k+1}^{n_{k+1}}{|a_n|^2}>1##. Then of course, we'd like to pick b's in such a way that ##\sum{a_n b_n}##, while they still converge in the square. But since we don't know how far apart the ##n_k## are, I can't figure out a way to do that. If ##b_n=\left|\frac{\bar{a_n}}{n}\right|##, then the b's converge the way we want them to, but it's not clear that ##\sum{a_n b_n}## diverges. On the other hand, if we choose something that depends on ##a_n## in some way, which seems more promising in some ways, then it's not clear that the b's converge the way they're supposed to. For instance if ##b_n=\left|\frac{\bar{a_n}}{c_n}\right|##, where ##c_n=k## when ##k<n\leq k+1##, then the c's are growing pretty slowly (presumably) so I shouldn't think the b's would converge properly.
     
    Last edited: Aug 23, 2012
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  3. Aug 23, 2012 #2

    Bacle2

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    It seems like it may have to see with Cauchy-Schwarz:

    We know if (bn,bn) < oo , then (an,bn)^2<oo . Let me try some more.
     
  4. Aug 23, 2012 #3

    vela

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    This argument doesn't work for an=1 since ##\sum{|b_n|^2}## will diverge.
     
  5. Aug 23, 2012 #4
    .....##\sum{\frac{1}{n^2}}## converges... doesn't it?
     
  6. Aug 23, 2012 #5

    vela

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    D'oh! Never mind. ;)
     
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