Suppose we want to test the following series for convergence
$$\sum_{n=1}^{\infty}\Big(1-n \sin(n^{-1})\Big)$$
I have found a solution that uses the asymptotic comparison test and uses the following equivalence: $$ \sin(n^{-1})\sim {1\over n} - {1\over6n^3};$$ $$1-n \sin(n^{-1}) \sim {1\over6n^2}$$ So the series converges absolutely because ${1\over6n^2}$ converges absolutely.
My question is, why choose ${1\over n} - {1\over6n^3}$ and not some different equivalence? If we used ${1\over n} - {1\over n^2}$, the resulting series would have been:
$$\sin(n^{-1})\sim {1\over n} - {1\over n^2};$$
$$ 1-n \sin(n^{-1}) \sim {1\over n},$$ which diverges and so the result would be incorrect. Why is that?