# Simple Real Analysis Problem - Using comparison test to prove a series diverges.

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?

• Why doesn't $a_nb_n$ converge to $0$? Dec 4 '19 at 6:39
• @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? Dec 4 '19 at 6:43
• Take $a_n=1/n$ and $b_n=1$ for an example. Dec 4 '19 at 6:46
• @EclipseSun True... Dec 4 '19 at 6:49

Hint:

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

• What can you say about $\alpha$? How can we deduce $a_n < a_nb_n$ from this? Dec 4 '19 at 6:53
• 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.
– RRL
Dec 4 '19 at 7:01
• Can you say the same even when $a_n=1/n$, the harmonic series? Dec 4 '19 at 7:04
• I just realized it doesn't matter since we are given that $\sum a_n$ diverges Dec 4 '19 at 7:05
• Yes — $\sum \frac{c}{n} =c\sum \frac{1}{n} = \infty$
– RRL
Dec 4 '19 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.

• I understand. But what can we know about $b$? How can we tell $a_n < a_nb_n$? Dec 4 '19 at 6:50