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

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

• $$\lim_{n\to\infty}\arcsin\dfrac k{n^2}\approx\lim\dfrac k{n^2}$$ – lab bhattacharjee Jun 4 '19 at 18:24
• @labbhattacharjee Thank you a lot! – Rareș Stanca Jun 4 '19 at 18:32

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.