Alisha Pitts

2022-01-24

Compute
$\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\mathrm{arcsin}\left(\frac{k}{{n}^{2}}\right)$

Mazzuranavf

Expert

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

ebbonxah

Expert

Another way to look at this is to observe that
$\mathrm{arcsin}\left(\frac{k}{{n}^{2}}\right)=\frac{k}{{n}^{2}}{\int }_{0}^{1}\frac{du}{\sqrt{1-\frac{{k}^{2}}{{n}^{4}}{u}^{2}}}$
Then you can reverse order of summation and integration and get that the sum equals
${\int }_{0}^{1}du\frac{1}{n}\sum _{k=1}^{n}\frac{\left(\frac{k}{n}\right)}{\sqrt{1-\frac{{k}^{2}}{{n}^{4}}{u}^{2}}}$
We almost have a Riemann sum, but not quite. The good news is that we can convert this to a Riemann sum by subbing $u=nv$ in the integral. The result is
$n{\int }_{0}^{\frac{1}{n}}dv\frac{1}{n}\sum _{k=1}^{n}\frac{\left(\frac{k}{n}\right)}{\sqrt{1-\frac{{k}^{2}}{{n}^{2}}{v}^{2}}}$
Now we have a Riemann sum, and as $n\to \mathrm{\infty }$ it becomes the integral
${\int }_{0}^{1}dx\frac{x}{\sqrt{1-{v}^{2}{x}^{2}}}=\frac{1-\sqrt{1-{v}^{2}}}{{v}^{2}}$
The limit we seek is then
$\underset{n\to \mathrm{\infty }}{lim}\left(n{\int }_{0}^{\frac{1}{n}}dv\frac{1-\sqrt{1-{v}^{2}}}{{v}^{2}}\right)=\underset{n\to \mathrm{\infty }}{lim}\left(n{\int }_{0}^{\frac{\mathrm{arcsin}1}{n}}d\theta \frac{\mathrm{cos}\theta }{1+\mathrm{cos}\theta }\right)$
which 1/2.

RizerMix

Expert

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 $\mathrm{sin}\theta \approx \theta$ so likewise $\mathrm{arcsin}x\approx x$ for small x. This means that $\sum _{k=1}^{n}\mathrm{arcsin}\left(\frac{k}{{n}^{2}}\right)\approx \sum _{k=1}^{n}\frac{k}{{n}^{2}}=\frac{1}{{n}^{2}}\sum _{k=1}^{n}k$ $=\frac{n\left(n+1\right)}{2{n}^{2}}$