Limit comparison test for series

I'm confused by one thing

lct states that if you have two series, $$a_n$$ and $$b_n$$ if taking the limit of two (where $$bn$$ is the decent comparison where it's positive for all n meaning it's limit doesn't equal 0) then both series behave the same way, ie both converge or diverge.

If $$L = 0$$ (which you can only have if $$a_n$$'s limit is 0 as the limit of $$b_n$$ can't be zero) and you know $$b_n$$ converges then $$a_n$$ must also converge, this I understand due to the vanishing condition, you can only get 0 if $$a_n$$'s limit is 0.

What is confusing me this, if $$L=\infty$$ and you know $$b_n$$ diverges, how does this imply $$b_n$$ also diverges? You would have an instance of an indeterminate form right where it's $$\frac{\infty}{\infty}$$.

• The part where you can only get 0 if $a_n$ is zero is a consequence of the series $a_n$ being convergent right? I think I'm not applying that same it's a consequence logic to the $L=\infty$ scenario
– Krio
Aug 21 '19 at 4:01
• Are you given two sequences $a_n$ and $b_n$ or are you given two series $\sum a_n$ and $\sum b_n$? These are totally different things. By not expressing clearly what you're talking about you confuse yourself (not to mention that your question is a bit difficult to comprehend). Aug 22 '19 at 4:26

The limit comparison states that for two series $$\Sigma_n a_n$$ and $$\Sigma_n b_n$$ with $$a_n\geq 0, b_n > 0$$ for all $$n$$ we have

$$L=\lim_{n \to \infty} \frac{a_n}{b_n}$$

1. If $$0 (i.e., L is a positive, finite number) then either the series $$\Sigma_n a_n$$ and $$\Sigma_n b_n$$ both converge or both diverge.
2. If $$L=0$$ and $$\Sigma_n b_n$$ converges, then $$\Sigma_n a_n$$ converges.
3. If $$L=\infty$$ and $$\Sigma_n b_n$$ diverges, then $$\Sigma_n a_n$$ diverges.

The idea is that as $$L=\lim_{n \to \infty} \dfrac{a_n}{b_n}$$, eventually we will have $$a_n\approx Lb_n$$. So if one of the series converges (or diverges) so does the other since the two are essentially scalar multiples of each other.

In the second case, if

$$\lim_{n \to \infty} \dfrac{a_n}{b_n}=L=0$$

and we know that $$\Sigma_n b_n$$ converges, then $$\Sigma_n a_n$$ must also converge since $$a_n\approx Lb_n$$ and $$L=0$$. In the third case, if

$$\lim_{n \to \infty} \dfrac{a_n}{b_n}=L=\infty$$

and we know that $$\Sigma_n b_n$$ diverges, then $$\Sigma_n a_n$$ must also diverge since $$a_n\approx Lb_n$$ and $$L=\infty$$.