Show that if $\displaystyle \sum{a_n}$ is a convergent positive-term series, then the series $\displaystyle \sum{sin(a_n)}$ also converges.
but $\displaystyle \sin a_n$ is not necessarily positive! we can remove this difficulty by using the inequality $\displaystyle |\sin x| \leq |x|.$
so if $\displaystyle \sum a_n$ is convergent with $\displaystyle a_n \geq 0,$ then $\displaystyle \sum \sin a_n$ is absolutely convergent and hence convergent.
Actually $\displaystyle \sin(a_n)>0$ for almost all $\displaystyle n$.
It was given that $\displaystyle (a_n) $ is a positive sequence and we know that $\displaystyle (a_n)\to 0 $.
Therefore, for almost all $\displaystyle n$ we have $\displaystyle 0< a_n<\frac{\pi}{2}$ which implies $\displaystyle 0<\sin(a_n)<a_n$.
So you are mistaken. Comparison works.
P.S. If $\displaystyle 0<x$ then $\displaystyle \sin(x)<x$.