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I'm having trouble starting this proof. From the initial hypothesis, we know for some $n>N$, $b_n>0$. Since $a_n>0$ and $a_n$ diverges, for some $n>N'$, $a_n\geq \varepsilon$. Then $a_n b_n>0$, so if $\sum a_n b_n$ is a series of positive numbers. I'm not sure how to show it diverges however, so any help would be appreciated!

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  • $\begingroup$ @qbert isn't the answer below correct regardless? $\endgroup$ – clark Dec 3 '18 at 3:03
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Let $$ \liminf_{n\to \infty}b_n=2\delta>0 $$ Then, for large enough $N$, $b_n\geq \delta$ for all $n\geq N$. Then, $$ \sum_{n=N}^\infty b_na_n\geq \delta\sum_{n=N}^\infty a_n=+\infty $$ since $\lim_{n\to \infty}a_n\ne 0$.

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  • $\begingroup$ Thank you! My only question is, where does the 2 go in the $2\delta$ term? $\endgroup$ – t.perez Dec 5 '18 at 2:12
  • $\begingroup$ okay, that makes sense now. thank you! $\endgroup$ – t.perez Dec 5 '18 at 2:24

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