Answer
Converges absolutely
Work Step by Step
Let us consider $u_n=(-1)^n \dfrac{\sin n}{n^2}$ and $u_n$ refers to all positive values of $n$.
Also, $|u_n|=\dfrac{\sin n}{n^2}$
Apply the limit comparison test.
Thus, $ \Sigma_{n=1}^\infty \dfrac{\sin n}{n^2}\leq \Sigma_{n=1}^\infty\dfrac{1}{n^2}
=\Sigma_{n=1}^\infty\dfrac{1}{n^2}$ \
But $\Sigma_{n=1}^\infty\dfrac{1}{n^2}$ shows a convergent p-series, so the series converges.
Thus, the series Converges absolutely.