Convergence of the series of sin(n)/n

Step 1: the partial sums of $\sin k$ are bounded. Using $2\sin\frac12\sin k=\cos\left(k-\frac12\right)-\cos\left(k+\frac12\right)$ and telescoping,
$$B_n=\frac{\cos\frac12-\cos\left(n+\frac12\right)}{2\sin\frac12},\qquad|B_n|\le\frac1{\sin\frac12}=:M.$$

Step 2: Abel's summation by parts. With $a_k=\frac1k$ and $b_k=\sin k$,
$$\sum_{k=1}^na_kb_k=a_nB_n+\sum_{k=1}^{n-1}(a_k-a_{k+1})B_k.$$

Step 3: convergence. Since $a_n\to0$ and $|B_n|\le M$, $a_nB_n\to0$. The second sum converges absolutely because
$$\sum_{k}(a_k-a_{k+1})|B_k|\le M\sum_k(a_k-a_{k+1})=Ma_1<\infty$$
(the differences are positive and telescope). So the partial sums have a limit.

Remark. This is Dirichlet's test: *bounded partial sums times a monotone sequence tending to $0$ gives a convergent series*. The series is not absolutely convergent ($\sum\frac{|\sin n|}n=\infty$).