Show that the sequence is bounded?

Show that the sequence $\frac{3n + sin^2n}{6n}_{n \in \mathbb{N}}$ is bounded.

I know that $ 0 \leq sin^2n \leq 1$, so $\frac{3n}{6n} = \frac{1}{2} \leq \frac{3n + sin^2n}{6n} \leq \frac{3n + 1}{6n}$

So it is easy to see that it is bounded below by $\frac{1}{2}$, but I can't figure out how to show it is bounded above.