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Assume that $\sum a_n$ is convergent and $\sum b_n$ is absolutely convergent.To show that $\sum a_nb_n$ is absolutely convergent

My try::Consider the sequence of partial sums of $\sum |a_nb_n|$

$S_n=|a_1b_1|+|a_2b_2|+......+|a_nb_n|$

Now $(a_1b_1+a_2b_2+......+a_nb_n)\leq ( (a_1^2+a_2^2+..+a_n^2))^\frac{1}{2}((b_1^2+b_2^2+..+b_n^2)^\frac{1}{2})$

How to proceed from here?

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  • $\begingroup$ A weaker statement is true. If $(a_n)$ is bounded and $(b_n)$ is absolutely summable, $(a_nb_n)$ is absolutely summable. Can you prove this by comparison? $\endgroup$ – Pedro Tamaroff Jan 19 '15 at 6:52
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Hint: for large enough $n$, $|a_n|\leq 1$, say for $n\geq N$... So $$\sum_{n=0}^{N+p}|a_nb_n|\leq \underbrace{\sum_{n=0}^{N-1}|a_nb_n|}_{C}+\sum_{n=N}^{N+p}|b_n|\leq C+\sum_{n=0}^{+\infty}|b_n|\,.$$

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Hint: The sequence $(a_n)$ converges to $0$ and is bounded. Show that partial sums of $\sum a_nb_n$ form a Cauchy sequence.

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