Doubt on the comparison test: can I still evaluate $\lim \limits_{n \to \infty} \frac{a_{n}}{b_{n}}$ if $b_{n}$ might be $0$ for some $n$? Suppose that $(b_{n})_{n \in \mathbb{N}}$ is a sequence which is not identically equal to $0$ (but which may have elements equal to $0$). Suppose also that I know that $\sum b_{n}$ converges absolutely.
If I want to verify if another series $\sum |a_{n}|$ converges, can I use the comparison test, that is:$$\lim \limits_{n \to \infty} \frac{|a_{n}|}{|b_{n}|} = c \neq 0 \implies \sum |a_{n}|$$ converges? The problem here is that I might have $0$ in the denominator...
 A: In short, you cannot apply the limit comparison test if $b_n = 0$ for very many $n$. There are a few conditions when you can, however.


*

*If $b_n$ is eventually nonzero (i.e. if there is some $N > 0$ so that $b_n \neq 0$ for all $n > N$), then you can apply limit comparison. This is because the first finitely many terms don't contribute to the eventual convergence or divergence of the sums.

*If $a_n = 0$ whenever $b_n = 0$, then you can apply limit comparison. Morally, you can think of these terms of just never having been there at all. I have a hard time imagining non-contrived examples of this behaviour.

*As André Nicolas mentioned in a comment: split $\mathbb{N}$ into $\{ m : b_m = 0\}$ and $\{ n : b_n \neq 0 \}$. Then if the limit
$$ \lim \frac{\lvert a_n \rvert}{\lvert b_n \rvert} = c \neq 0$$
converges and the sum
$$ \sum_{m: b_m = 0} \lvert a_m \rvert < \infty,$$
converges then the series associated to $a_i$ converges if and only if the series associated to $b_i$ converges.
It might also be worth noting that one can also say things when the limit is $0$, as this indicates that the series associated to $|a_i|$ is smaller in magnitude then the series associated to $|b_i|$.
