The series converges.
Because | sin(x) | 1 for all real number x, we have that
1 / 2 n sin(n) | | 1 / 2 n |But is a special case of the geometric series, which converges. Hence, by the comparison test, the original series converges absolutely.
Note that, formally, we could say: Since
| 1 / 2 n sin(n) | | 1 / 2 n |for all n. We also have
| | <"Therefore", the series converges absolutely.