Show that it’s absolutely convergent, i.e., that $\sum_{n=1}^\infty\left|\frac{\sin na}{n^2}\right|$ converges. You should have no trouble doing this by comparing it with a simple series that you know converges.
Note, by the way, that it makes no sense at all to write $\lim_{n\to\infty}\frac{\sin na}{n^2}=\frac{[-1,1]}\infty\;:$ $\infty$ isn’t something by which you can divide, and $[-1,1]$ isn’t even a number. What you mean is this:
For all $n$, $\frac{-1}{n^2}\le\frac{\sin na}{n^2}\le\frac1{n^2}\;,$ so $0=\lim_{n\to\infty}\frac{-1}{n^2}\le\lim_{n\to\infty}\frac{\sin na}{n^2}\le\lim_{n\to\infty}\frac1{n^2}=0\;,$ and therefore by the squeeze theorem $\lim_{n\to\infty}\frac{\sin na}{n^2}=0\;.$