$$\sum\limits_{n=1}^{\infty}\frac{\sin n}{n}(1+\frac{1}{2}+\dots+\frac{1}{n})$$
Distributing we have$$\sum\limits_{n=1}^{\infty}\Bigl(\frac{\sin n}{n}+\frac{\sin n}{2n}+\dots+\frac{\sin n}{n^2}\Bigr)$$The n-th partial sum of the series is$$ S_n = (\sin 1) + (\frac{\sin 2}{2} + \frac{\sin 2}{4}) + (\frac{\sin 3}{3} + \frac{\sin 3}{6} + \frac{\sin 3}{9}) + \dots + (\frac{\sin n}{n} + \dots + \frac{\sin n}{n^2})$$
$$\require{cancel}\xcancel{= \underbrace{n(\sin1)}_{\to \infty \text{ as } n\to \infty} + (n-1)\frac{\sin 2}{4}+\dots+2\cdot\frac{\sin (n-1)}{n(n-1)}+\underbrace{1\cdot\frac{\sin n}{n^2}}_{\to 0 \text{ as } n\to \infty}}$$
So as $n \to \infty$ the "first" terms approach $\infty$ and the "last" terms approach $0$, therefore the series diverges. But I'm not sure how to properly state this, which makes me wonder whether there's a more a straightforward way to solve this.