As indicated by Todd Trimble in comments, we can use the Dirichlet test; here, since $$\sin(n)\sin(n^2)=\frac12\big( \cos n(n-1) - \cos n(n+1) \big)$$ we have a telescopic sum $$\sum_{n=1}^M \sin(n)\sin(n^2)=\frac12-\frac12 \cos M(M+1),$$ that does not exceed $1$ in absolute value.

As indicated by Todd Trimble in comments, we can use the Dirichlet test; here, since $$\sin(n)\sin(n^2)=\frac12\big( \cos n(n-1) - \cos n(n+1) \big)$$ we have a telescopic sum $$\sum_{n=1}^M \sin(n)\sin(n^2)=\frac12-\frac12 \cos M(M+1)=\sin^2\Big(\frac{M(M+1)}2\Big),$$ that does not exceed $1$ in absolute value.