How to evaluate $\lim_{n \to \infty} \sum_{i=1}^n \arcsin{(u \cdot i/n)}$ without Riemann-sums

Question

How to evaluate the limit $$\lim_{n \to \infty} \sum_{i=1}^n \dfrac{1}{n}\arcsin{(u \cdot i/n)}$$ Where $0 < u \leq 1$. There is a closed form expression, but I just don't know how to find it.

Without Riemann-summs.

Answer 1

Note that

$$\dfrac{1}{n}\arcsin{(u \cdot i/n)}=i\cdot \left(\frac{u}{n^2}+o\left(\frac1{n^2}\right)\right)$$

then

$$\sum_{i=1}^n \dfrac{1}{n}\arcsin{(u \cdot i/n)}=\left(\frac{u}{n^2}+o\left(\frac1{n^2}\right)\right)\sum_{i=1}^n i =\frac{u}{n^2}\frac{n(n+1)}2+o\left(\frac1{n^2}\right)\frac{n(n+1)}2=$$

$$=\frac{u}{2}\frac{n(n+1)}{n^2}+\frac{o\left(\frac1{n^2}\right)}{\frac1{n^2}}\frac{n(n+1)}{2n^2}\to \frac u 2$$