Limit comparison test for series

Question

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 $\frac{\infty}{\infty}$.

Answer 1

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 $0<L<\infty$ (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 $L=\infty$.