A few people have asked about the limit comparison test during office hours. It’s stated on page 564 of your text, but we can reword it in a way that might make it easier to understand.
First, imagine that instead of assuming that 
, we assumed that 
for all values of n, not just in the limit.
If 
, we could write 
and 
. Either way, 
is a multiple of 
and vice versa. Multiplying a series by a constant (non-zero, non-infinite) doesn’t change whether it converges or diverges. Therefore, either both series 
and 
converge or they both diverge….both series have to behave the same way.
If 
is very very large, then 
, that is, is a tiny multiple of . Therefore, if you know that converges, then must also converge.
If is very very small, then , that is, is a tiny multiple of . Therefore, if you know that converges, then must also converge.
None of what I’ve said so far should sound too crazy. What’s interesting about the limit comparison test is that you don’t have to assume that for all values of 
, you just have to assume that 
is close to only for large values of . If , what this says is that is eventually almost a multiple of and vice versa. Therefore, either both series and converge or they both diverge…both series have to behave the same way.
Additionally, if you let actually approach 
, then is eventually almost a tiny multiple of . Therefore, if you know that converges, then must also converge.
Finally, if you let actually approach 
, then is eventually almost a tiny multiple of . Therefore, if you know that converges, then must also converge.
tl;dr: If the individual terms of one series are eventually almost equal to a multiple of the individual terms of another series, then both series have to behave the same way (both converge or both diverge).
