Cancellation revealed by a Taylor expansion

Idea. The two terms $\frac1n$ and $\sin\frac1n$ almost cancel, so comparing the difference with $\frac1n$ would be a blunder.

Computation.
$$\frac1n-\sin\frac1n=\frac1n-\left(\frac1n-\frac1{6n^3}+o\!\left(\frac1{n^3}\right)\right)=\frac1{6n^3}+o\!\left(\frac1{n^3}\right).$$
So the terms are positive for large $n$ and equivalent to $\frac1{6n^3}$.

Conclusion. $\sum\frac1{n^3}$ converges, so the serie $\displaystyle\sum_{n\ge1}\left(\frac1n-\sin\frac1n\right)$ converges.

Remark. Each time the leading terms cancel, one has to go one order further in the expansion.