I must determine whether if the following series converges, converges absolutely, or diverges: $$\sum_{n=1}^\infty\sin(n)\sin\left(\frac{\pi}{2n}\right)$$ By the comparison test, I have already found that $\sum\limits_{n=1}^\infty \left(\sin\left(\frac{\pi}{2n}\right)\right)^p$ converges for $p>1$ and diverges for $p \leq 1$. Thus, $ \sum\limits_{n=1}^\infty\sin\left(\frac{\pi}{2n}\right)$ diverges by this criterion. I suspect the entire series will also diverge, and that I have to use the comparison test, but I encountered an issue:
Since $-1 \leq \sin n \leq 1$, we have that $\sin(n)\sin\left(\frac{\pi}{2n}\right) \leq \sin\left(\frac{\pi}{2n}\right)$. This would be useful if the series represented by the term on the right converged; in its current state, this cannot be used to prove divergence.
Is my reasoning wrong? Should I be using another test for this series? Thank you.