n^2 - 1 / (n^3 + 6n)
If I use the nth divergence test, I plug ∞ in (limit as n -> ∞) for n and since the degree on the bottom is larger I get 0, which means it converges.
However, if I use the limit comparison test and compare it to: n^2/n^3, which = 1/n, which diverges -> n^2 - 1 / (n^3 + 6n)...