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.