Let $a_n > 0$ and suppose that $\sum a_n$ diverges. Prove that $\sum a_n b_n$ diverges for all sequences $\{b_n\}_n$ with $\liminf_n b_n >0$.

I know this is a simple problem. I already proved using the fact that $a_n b_n$ does not converge to 0 and thus the series must diverge. However, I am not sure how to prove this using the comparison test with $\sum a_n$. We would need to deduce that $a_n < a_nb_n$ but how can we do that? Does $\liminf_n b_n >0$ implies this? What if $b_n=0.01$ for all n?

  • 1
    $\begingroup$ Why doesn't $a_nb_n$ converge to $0$? $\endgroup$ Dec 4, 2019 at 6:39
  • $\begingroup$ @EclipseSun Since ∑𝑎𝑛 diverges, $a_n$ must diverge or converge to a non-zero number. And since $b_n$>0 for all n, $a_nb_n$ doesn't converge to 0. This was my logic. Do you think this is incorrect? $\endgroup$ Dec 4, 2019 at 6:43
  • $\begingroup$ Take $a_n=1/n$ and $b_n=1$ for an example. $\endgroup$ Dec 4, 2019 at 6:46
  • $\begingroup$ @EclipseSun True... $\endgroup$ Dec 4, 2019 at 6:49

2 Answers 2



If $\liminf b_n = \alpha >0 $ then there exists $N$ such that $ b_n > \alpha/2$ for all $n >N$

  • $\begingroup$ What can you say about $\alpha$? How can we deduce $a_n < a_nb_n$ from this? $\endgroup$ Dec 4, 2019 at 6:53
  • $\begingroup$ With $a_n>0$, $a_nb_n$ must be greater than a constant $c= \alpha/2$ times $a_n$ for all sufficiently large $n$ and so $\sum a_n b_n$ diverges by the comparison test since $\sum ca_n$ diverges. $\endgroup$
    – RRL
    Dec 4, 2019 at 7:01
  • $\begingroup$ Can you say the same even when $a_n=1/n$, the harmonic series? $\endgroup$ Dec 4, 2019 at 7:04
  • $\begingroup$ I just realized it doesn't matter since we are given that $\sum a_n$ diverges $\endgroup$ Dec 4, 2019 at 7:05
  • $\begingroup$ Yes — $\sum \frac{c}{n} =c\sum \frac{1}{n} = \infty$ $\endgroup$
    – RRL
    Dec 4, 2019 at 7:07

First few terms of a series do not affect convergence or divergence of the series. There exist $b >0$ and $m$ such that $b_n \geq b$ for all $n \geq m$. Hence $\sum a_nb_n$ is divergent.

  • $\begingroup$ I understand. But what can we know about $b$? How can we tell $a_n < a_nb_n$? $\endgroup$ Dec 4, 2019 at 6:50

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