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.

Another way to look at this is to observe that
$$\arcsin{\left ( \frac{k}{n^2} \right )} = \frac{k}{n^2} \int_0^1 \frac{du}{\sqrt{1-\frac{k^2}{n^4} u^2}} $$
Then you can reverse order of summation and integration and get that the sum equals
$$\int_0^1 du \, \frac1{n} \sum_{k=1}^n \frac{(k/n)}{\sqrt{1-\frac{k^2}{n^4} u^2}} $$
We almost have a Riemann sum, but not quite.  The good news is that we can convert this to a Riemann sum by subbing $u=n v$ in the integral.  The result is
$$n \int_0^{1/n} dv \, \frac1{n} \sum_{k=1}^n \frac{(k/n)}{\sqrt{1-\frac{k^2}{n^2} v^2}} $$
Now we have a Riemann sum, and as $n \to \infty$ it becomes the integral
$$\int_0^1 dx \, \frac{x}{\sqrt{1-v^2 x^2}} = \frac{1-\sqrt{1-v^2}}{v^2} $$
The limit we seek is then
$$\lim_{n \to \infty} \left (n \int_0^{1/n} dv \, \frac{1-\sqrt{1-v^2}}{v^2} \right ) = \lim_{n \to \infty} \left (n \int_0^{\arcsin{1/n}} d\theta \, \frac{\cos{\theta}}{1+\cos{\theta}} \right )$$
which is $1/2$.
