Evaluate $\lim_{n\rightarrow\infty}\sum_{k=1}^n\arcsin(\frac k{n^2})$ 
Compute $$\lim_{n\to\infty}\sum_{k=1}^n\arcsin\left(\frac k{n^2}\right)$$

Hello, I'm deeply sorry but I don't know how to approach any infinite sum that involves $\arcsin$, so I couldn't do anything to this question. Any hints/tips would be appreciated. I know I have to make it somehow telescopic but I don't know how to use formulas like 
$$\arcsin x-\arcsin y=\arcsin\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)$$
My knowledge level is 12th grade.
I tried to put it in between $$\arcsin\frac 1{n^2}< \sum_{k=1}^n\arcsin\frac k{n^2} <\arcsin\frac n{n^2}$$ so then $L=0$, but it's wrong.

Here is a nonrigorous approach that can be filled out to a solution:
Observe that each $\frac{k}{n^2}$ is going to be pretty small once $n$ is large. For small angles $\theta$ you have that $\sin \theta \approx \theta$ so likewise $\arcsin x \approx x$ for small $x$. This means that
$$\sum_{k=1}^n \arcsin \left( \frac k{n^2} \right) \approx \sum_{k=1}^n \frac k{n^2} = \frac 1{n^2} \sum_{k=1}^n k = \frac{n(n+1)}{2n^2}.$$
What happens as $n \to \infty$?
