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If $\sum (a_n)^2$ converges and $\sum (b_n)^2$ converges, does $\sum (a_n)(b_n)$ converge?

Could someone help me to solve this or at least give me a hint?, I have tried using Cauchy's criterion, the Dirichlet test for convergence, etc, but I can´t prove it.Honestly I don´t know where to start. Any help will be appreciated.

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    $\begingroup$ Hint: Cauchy-Schwartz inequality. $\endgroup$
    – Jose27
    May 13, 2015 at 2:31
  • $\begingroup$ using cauchy will require knowing either 1) $l^2$ is an inner product space or 2) monotone convergence for real numbers, which seems slightly complicated for such a simple problem (not to say I don't approve!) $\endgroup$
    – cats
    May 13, 2015 at 2:36
  • $\begingroup$ Vote to close. The OP has not returned for 2 years. $\endgroup$
    – GEdgar
    Apr 28, 2021 at 17:16
  • $\begingroup$ @cats how is monotone convergence complicated for any problem? Without there's no comparison test which is the simplest of the convergence tests $\endgroup$ Feb 22, 2023 at 9:09

1 Answer 1

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Start from here :$$(|a_n|-|b_n|)^2=a_n^2+b_n^2-2|a_nb_n|\ge 0$$

$$\implies |a_nb_n|\le \frac{1}{2}(a_n^2+b_n^2)$$By comparison test, $\sum a_nb_n$ is absolutely convergent , hence convergent.

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  • $\begingroup$ But how do I know that $(a_n)(b_n)\ge0$ so that I can apply the comparison test? Can I really try sequences just like numbers? thanks $\endgroup$
    – mobzopi
    May 13, 2015 at 2:36
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    $\begingroup$ Easy enough fix, but you might need to specify that $a_n$ and $b_n$ are positive or else take the absolute value and then apply the above result to show it absolutely converges and therefore convereges $\endgroup$
    – CPM
    May 13, 2015 at 2:37
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    $\begingroup$ Thanks, it was so easy but I didn't see it... $\endgroup$
    – mobzopi
    May 13, 2015 at 2:47
  • $\begingroup$ would this work for complex numbers? $\endgroup$
    – Nimish
    Feb 26, 2020 at 2:08
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    $\begingroup$ @Nimish absolute convergence implies convergence $\endgroup$ Mar 5, 2020 at 13:50

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