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

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

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  • $\begingroup$ Thank you a whole freaking lot!! Do you have any idea how is this method called? $\endgroup$ – Rareș Stanca Jun 4 '19 at 18:31
  • $\begingroup$ @rares_st I don't think it's special enough to have a name. $\endgroup$ – Angina Seng Jun 4 '19 at 18:32

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