Show that the sequence is bounded?

Question

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.

Answer 1

$\frac {3n+1}{6n}=\frac {3 +1/n}{6} < \frac 4 6 = 2/3$

Answer 2

$\frac{3n}{6n} = \frac{1}{2} \leq \frac{3n + \sin^2n}{6n} \leq \frac{3n + 1}{6n}$

$\frac{1}{2} \leq \frac{3n + \sin^2n}{6n} \leq \frac{3n}{6n}+\frac{1}{6n}$

$\frac{1}{2} \leq \frac{3n + \sin^2n}{6n} \leq \frac{1}{2}+\frac{1}{6n}\leq\frac{1}{2}+\frac{1}{6}$

In the last step I used the property that $1/6n$ is decreasing for all $n$ and hence has a maximum for $n=1$.