How to find $\lim_{n\to\infty}\sum_{k=1}^n \arcsin\frac{k}{n^2}$?

Question

$$\lim_{n\to\infty}\sum_{k=1}^n \arcsin\frac{k}{n^2}$$ Any hints on how to approach this problem in the first place? The answer should be $\frac12$.

I tried transforming it so I can use a Riemann sum, but it doesn't seem correct.

Answer 1

Following lab bhattacharjee's hint, $\arcsin x=x+O(x^3)$ for small $x$. Therefore $$\sum_{k=1}^n\arcsin\frac k{n^2}=\sum_{k=1}^n \frac k{n^2}+O\left(\sum_{k=1}^n \frac {k^3}{n^6}\right).$$ In that second sum, the largest term is $1/n^3$ and there are $n$ terms so that $$\sum_{k=1}^n\arcsin\frac k{n^2}=\sum_{k=1}^n \frac k{n^2}+O\left(\frac1{n^2}\right)$$ etc.