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Suppose {$a_n$} and {$b_n$} are two sequences of positive real numbers such that $\displaystyle \sum_{n=1}^{\infty} a_n^2 < \infty$ and $\displaystyle \sum_{n=1}^{\infty} b_n^2 < \infty$ Prove that $\displaystyle \sum_{n=1}^{\infty} a_nb_n$, $\displaystyle \sum_{n=1}^{\infty} (a_n + b_n)^2$, $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n}$ are convergent.

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Have you heard of the Cauchy Schwarz Inequality? – Alex R. Nov 7 '12 at 6:48

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Since $(a_n-b_n)^2=a_n^2-2a_nb_n+b_n^2\ge0$, we have $a_nb_n\le(a_n^2+b_n^2)/2$. Thus $\sum a_nb_n$ is dominated by a convergent series. The convergence of $\sum(a_n+b_n)^2=\sum(a_n^2+2a_nb_n+b_n^2)$ follows immediately. For the last one, note that by the Cauchy–Schwarz inequality

$$ \sum_{n=1}^m\frac{a_n}n\le\sqrt{\sum_{n=1}^ma_n^2\sum_{n=1}^m\frac1{n^2}}\le\sqrt{\sum_{n=1}^\infty a_n^2\sum_{n=1}^\infty\frac1{n^2}}\lt\infty\;. $$

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What if not all of the terms in {$a_n$} and {$b_n$} were positive and you had to prove convergence of $\sum_{n=1}^{\infty} |a_nb_n|$? – hallaplay835 Jan 28 '14 at 20:51

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