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


*

*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.

*If $L=0$ and $\Sigma_n b_n$ converges, then $\Sigma_n a_n$ converges.

*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 $L=\infty$.
