The value of the series: $\sum_{p=1}^{n} \sin(\frac{p}{n^2})$ 
Find the limit as $n\rightarrow \infty$ of:
$$\sum_{p=1}^{n} \sin\left(\frac{p}{n^2}\right)$$

This question seems odd to me because it was included in my differentiability problems set.
My Attempt:
We have:  $\sin(x) \le x $, Thus clearly:
$$\sum_{p=1}^{n} \sin\left(\frac{p}{n^2}\right) \le \sum_{p=1}^{n} \frac{p}{n^2}$$
I have to prove that:
$$\sum_{p=1}^{n} \frac{p}{n^2} \rightarrow \frac12$$
And:
$$\frac{1}{2}\le\sum_{p=1}^{n} \sin\left(\frac{p}{n^2}\right)$$
To conclude that the series converges to $\frac12$.
Though, I think there exists another solution that uses derivative or something similar.
Update:
I proved by double summation that:
$$\sum_{p=1}^{n} \frac{p}{n^2} = \frac{1+\frac1n}{2}$$
So it clearly goes to $\frac12$
 A: Hint:
It can be done with plain trigonometry — there is a high-school formula for the sum of sines of arcs in arithmetic progression:
$$\sin a + \sin(a+\theta)+ \sin(a+2\theta)+ \dots + \sin(a+n\theta) = \frac{\sin\cfrac{(n+1)\theta}{2}}{\sin\cfrac{\theta}{2}}\,\sin\Bigl(a+\frac{n\theta}{2}\Bigr)$$
A: Maple says
$$
\sum_{p=1}^n\sin\left(\frac{p}{n^2}\right) =
{\frac {\sin \left( \frac{1}{n} \right) \cos \left( \frac{1}{n^2} \right) +
 \left( \cos \left( \frac{1}{n} \right) -1 \right) \sin \left( \frac{1}{n^2}
 \right) -\sin \left( \frac{1}{n} \right) }{2\,\cos \left( \frac{1}{n^2}
 \right) -2}}
$$
A: Starting from @GEdgar's answer, a simpler formula could be
$$S_n=\sum_{p=1}^n\sin\left(\frac{p}{n^2}\right) =\frac{1}{2} \left(\sin \left(\frac{1}{n}\right)-\left(\cos
   \left(\frac{1}{n}\right)-1\right) \cot \left(\frac{1}{2 n^2}\right)\right)$$ and, if $n$ is large
$$S_n=\frac{1}{2}+\frac{1}{2 n}-\frac{1}{24 n^2}-\frac{1}{12
   n^3}+O\left(\frac{1}{n^4}\right)$$
For $n=10$, the "exact" value is $0.549496$ while the above truncated expression gives $\frac{1099}{2000}=0.549500$.
