Answer
Absolutely convergent.
Work Step by Step
We test the series for absolute convergence by direct comparison.
Since $\left| {\sin k} \right| \le 1$, it follows that
$\mathop \sum \limits_{k = 1}^\infty \left| {\dfrac{{\sin k}}{{{k^3}}}} \right| \le \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^3}}}$
Since the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^3}}}$ is a convergent $p$-series ($p = 3 \gt 1$), by the Direct Comparison Test, the series $\mathop \sum \limits_{k = 1}^\infty \left| {\dfrac{{\sin k}}{{{k^3}}}} \right|$ converges. Therefore, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{{\sin k}}{{{k^3}}}$ converges absolutely.
