Alisha Pitts

Answered

2022-01-24

Compute

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

Answer & Explanation

Mazzuranavf

Expert

2022-01-25Added 10 answers

Recall for any $\theta \in (0,\frac{\pi}{2})$ , we have the inequality

$\mathrm{sin}\theta <\theta <\mathrm{tan}\theta$

This means for any$x\in (0,1)$ , 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

$\underset{n\to \mathrm{\infty}}{lim}\sum _{k=1}^{n}\mathrm{arcsin}\frac{k}{{n}^{2}}=\frac{1}{2}$

This means for any

For any

As

ebbonxah

Expert

2022-01-26Added 15 answers

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.

Then you can reverse order of summation and integration and get that the sum equals

We almost have a Riemann sum, but not quite. The good news is that we can convert this to a Riemann sum by subbing

Now we have a Riemann sum, and as

The limit we seek is then

which 1/2.

RizerMix

Expert

2022-01-27Added 437 answers

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

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