Does $a_n = \sin (\tfrac{\pi}{2}n + \tfrac{\pi}{3})\sqrt{n}$ converge? The question sounds like:

Let $a_n = \sin (\frac{\pi}{2}n + \frac{\pi}{3})n^{0.5}$ for $n = 1, 2, 3,...$
Check if $\{a_n\}$ from $n = 1$ to infinity is convergent or divergent.

So far, I have the following:
$$\{a_n\} = \sin \left(n\frac{\pi}{2} + \frac{\pi}{3}\right)\sqrt{n}$$
$$= \lim\limits_{n\to \infty} \sin \left(n\frac{\pi}{2}\right)\cdot\lim\limits_ {n\to \infty} \left(\frac{\pi}{3}\right) \sqrt{n}$$
I know that the limit of $\sin$ is bounded between $-1$ and $1$. Am I on the right path?

I assume your sequence is $\{a_n\}$ where $a_n = \sin(\frac{\pi n}{2} + \frac{\pi}{3}) \cdot \sqrt{n}$, for $n = 1, 2, 3, ...$. 
Using the angle sum formula, we see that 
$\sin(\frac{\pi n}{2} + \frac{\pi}{3}) = \cos(\frac{\pi n}{2}) \; \sin(\frac{\pi}{3}) + \sin(\frac{\pi n}{2})\; \cos(\frac{\pi}{3}) = $
$\frac{\sqrt{3}}{2}$ for $n = 4k$,
$ \frac{-\sqrt{3}}{2}$ for $n = 4k + 2$,
$ \frac{1}{2}$ for $n = 4k + 1$, and 
$\frac{-1}{2}$ for $n = 4k + 3$.
We have
$\lim_{k\to \infty} a_{4k} = \lim_{k \to \infty} a_{4k + 1} = \frac{\sqrt{3}}{2} \cdot(\lim_{k \to \infty} \sqrt{4k}) = \frac{1}{2} \cdot(\lim_{k \to \infty} \sqrt{4k+1}) = \infty$, 
and 
$\lim_{k \to \infty} a_{4k+2} = \lim_{k \to \infty} a_{4k + 3} = -\frac{\sqrt{3}}{2} \cdot(\lim_{k \to \infty} \sqrt{4k+2}) = -\frac{1}{2} \cdot (\lim_{k \to \infty} \sqrt{4k + 3}) = - \infty$.
So the limits of different index sequences are different. I believe we say that the limit does not exist. 
