The sine of an alternating term

Simplifying. Since $\sin$ is odd, $\sin\!\left(\frac{(-1)^n}n\right)=(-1)^n\sin\frac1n$.

Convergence. The sequence $b_n=\sin\frac1n$ is positive, decreasing (since $\frac1n$ decreases in $\left(0,\frac\pi2\right)$ where $\sin$ is increasing) and tends to $0$. By the alternating series test, the series converges.

Absolute convergence. $\left|\sin\frac{(-1)^n}n\right|=\sin\frac1n\sim\frac1n$, so $\sum\left|\cdot\right|$ diverges. The convergence is conditional.

Remark. The expansion route works too: $(-1)^n\left(\frac1n-\frac1{6n^3}+\dots\right)$ is a convergent alternating series plus an absolutely convergent one.