Compute limn→∞∑k=1narcsin⁡(kn2)

Recall for any ${\displaystyle \theta \in (0,\frac{\pi }{2})}$, we have the inequality
${\displaystyle \sin \theta <\theta <\tan \theta }$
This means for any ${\displaystyle x\in (0,1)}$, we have the bound
${\displaystyle x<\arcsin x<\frac{x}{\sqrt{1-x^2}}}$
For any ${\displaystyle n>1}$, this leads to
${\displaystyle \frac{n+1}{2n}=\sum _{k=1}^n\frac{k}{n^2}<\sum _{k=1}^n\arcsin \frac{k}{n^2}}$
${\displaystyle <\sum _{k=1}^n\frac{\frac{k}{n^2}}{\sqrt{1-{\left(\frac{k}{n^2}\right)}^2}}\le \frac{1}{\sqrt{1-\frac{1}{n^2}}}\sum _{k=1}^n\frac{k}{n^2}}$
${\displaystyle =\frac{n+1}{2\sqrt{n^2-1}}}$
As ${\displaystyle n\to \mathrm{\infty }}$, it is clear both sides converge to ${\displaystyle \frac{1}{2}}$. By squeezing, we obtain
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n\arcsin \frac{k}{n^2}=\frac{1}{2}}$

Another way to look at this is to observe that
${\displaystyle \arcsin \left(\frac{k}{n^2}\right)=\frac{k}{n^2}{\int }_0^1\frac{du}{\sqrt{1-\frac{k^2}{n^4}{u}^2}}}$
Then you can reverse order of summation and integration and get that the sum equals
${\displaystyle {\int }_0^{1}du\frac{1}{n}\sum _{k=1}^n\frac{\left(\frac{k}{n}\right)}{\sqrt{1-\frac{k^2}{n^4}{u}^2}}}$
We almost have a Riemann sum, but not quite. The good news is that we can convert this to a Riemann sum by subbing ${\displaystyle u=nv}$ in the integral. The result is
${\displaystyle n{\int }_0^{\frac{1}{n}}dv\frac{1}{n}\sum _{k=1}^n\frac{\left(\frac{k}{n}\right)}{\sqrt{1-\frac{k^2}{n^2}{v}^2}}}$
Now we have a Riemann sum, and as ${\displaystyle n\to \mathrm{\infty }}$ it becomes the integral
${\displaystyle {\int }_0^{1}dx\frac{x}{\sqrt{1-v^2{x}^2}}=\frac{1-\sqrt{1-v^2}}{v^2}}$
The limit we seek is then
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left(n{\int }_0^{\frac{1}{n}}dv\frac{1-\sqrt{1-v^2}}{v^2}\right)=\underset{n\to \mathrm{\infty }}{lim}\left(n{\int }_0^{\frac{\arcsin 1}{n}}d\theta \frac{\cos \theta }{1+\cos \theta }\right)}$
which 1/2.

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 $\sum _{k=1}^n\arcsin (\frac{k}{n^2})\approx \sum _{k=1}^n\frac{k}{n^2}=\frac{1}{n^2}\sum _{k=1}^{n}k$ $=\frac{n(n+1)}{2n^2}$