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Why I can't do:

$$\sum a_n b_n \le \sum a_n K$$

when $K$ is the bound of $b_n$ and then use the comparsion test, since $\sum a_n$ converges and therefore $\sum a_n K$ does too?

I've found Why is my proof wrong? (Rudin Ch 3 #8) If $\sum a_n$ converges, and if ${b_n}$ is monotonic and bounded, prove that $\sum a_n b_n$ converges. but mine uses the comparsion test, and this one uses epsilon delta definition

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That works if you know all the terms are nonnegative. If some $a_n$ might be negative, though, then your inequality may not be true, since $b_n\leq K$ does not necessarily imply $a_nb_n\leq a_n K$, and in any case such a term-by-term bound would not tell you $\sum a_nb_n$ converges if the terms aren't all nonnegative.

(Note that you very much need to use the assumption that $b_n$ is monotonic. For instance, if $a_n=(-1)^n/n$ and $b_n=(-1)^n$ then $\sum a_n$ converges and $(b_n)$ is bounded but $\sum a_nb_n$ does not converge.)

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