Convergence of three series via expansions

1. $\sin\frac1n=\frac1n-\frac1{6n^3}+o\!\left(\frac1{n^3}\right)$, so $\frac1n-\sin\frac1n\sim\frac1{6n^3}>0$. Convergent (comparison with $\sum\frac1{n^3}$).

2. By the expansion $\left(1+\frac1n\right)^n=e-\frac e{2n}+o\!\left(\frac1n\right)$, the term is $\sim\frac e{2n}>0$. Divergent (comparison with the harmonic series).

3. $n\sin\frac1n-1=-\frac1{6n^2}+o\!\left(\frac1{n^2}\right)\sim-\frac1{6n^2}$. The terms are eventually of constant sign (negative), so the series converges (absolutely) by comparison with $\sum\frac1{n^2}$.

Remark. The sign of the general term must be constant (at least eventually) for equivalents to give a valid conclusion. For terms of changing sign, equivalents may fail.