Evaluate $\lim_{n\rightarrow\infty}\sum_{k=1}^n\arcsin(\frac k{n^2})$

Question

Compute $$\lim_{n\to\infty}\sum_{k=1}^n\arcsin\left(\frac k{n^2}\right)$$

Hello, I'm deeply sorry but I don't know how to approach any infinite sum that involves $\arcsin$, so I couldn't do anything to this question. Any hints/tips would be appreciated. I know I have to make it somehow telescopic but I don't know how to use formulas like

$$\arcsin x-\arcsin y=\arcsin\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)$$

My knowledge level is 12th grade.

I tried to put it in between $$\arcsin\frac 1{n^2}< \sum_{k=1}^n\arcsin\frac k{n^2} <\arcsin\frac n{n^2}$$ so then $L=0$, but it's wrong.

Answer 1

Another way to look at this is to observe that

$$\arcsin{\left ( \frac{k}{n^2} \right )} = \frac{k}{n^2} \int_0^1 \frac{du}{\sqrt{1-\frac{k^2}{n^4} u^2}} $$

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

$$\int_0^1 du \, \frac1{n} \sum_{k=1}^n \frac{(k/n)}{\sqrt{1-\frac{k^2}{n^4} u^2}} $$

We almost have a Riemann sum, but not quite. The good news is that we can convert this to a Riemann sum by subbing $u=n v$ in the integral. The result is

$$n \int_0^{1/n} dv \, \frac1{n} \sum_{k=1}^n \frac{(k/n)}{\sqrt{1-\frac{k^2}{n^2} v^2}} $$

Now we have a Riemann sum, and as $n \to \infty$ 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

$$\lim_{n \to \infty} \left (n \int_0^{1/n} dv \, \frac{1-\sqrt{1-v^2}}{v^2} \right ) = \lim_{n \to \infty} \left (n \int_0^{\arcsin{1/n}} d\theta \, \frac{\cos{\theta}}{1+\cos{\theta}} \right )$$

which is $1/2$.

Answer 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 $$\sum_{k=1}^n \arcsin \left( \frac k{n^2} \right) \approx \sum_{k=1}^n \frac k{n^2} = \frac 1{n^2} \sum_{k=1}^n k = \frac{n(n+1)}{2n^2}.$$

What happens as $n \to \infty$?

Answer 3

Starting from the hint by Umberto P., to make things more rigorous, note that we have that for $n$ large

$$\arcsin\left(\frac k{n^2}\right)=\frac k {n^2}+o\left(\frac 1 {n^2}\right)$$

then

$$\sum_{k=1}^n \arcsin \left( \frac k{n^2} \right)=\sum_{k=1}^n \left(\frac k {n^2}+o\left(\frac 1 {n^2}\right)\right)=\frac{n(n+1)}{2n^2}+\frac{n(n+1)}{2}\cdot o\left(\frac 1 {n^2}\right)$$

now take the limit and recall that by definition

$$\frac{o\left(\frac 1 {n^2}\right)}{\frac 1 {n^2}}\to 0$$

Answer 4

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

This means for any $x \in (0,1)$, we have the bound $$x < \arcsin x < \frac{x}{\sqrt{1-x^2}}$$

For any $n > 1$, this leads to

$$\frac{n+1}{2n} = \sum_{k=1}^n \frac{k}{n^2} < \sum_{k=1}^n \arcsin\frac{k}{n^2} < \sum_{k=1}^n \frac{\frac{k}{n^2}}{\sqrt{1 - \left(\frac{k}{n^2}\right)^2}} \le \frac{1}{\sqrt{1 - \frac{1}{n^2}}}\sum_{k=1}^n\frac{k}{n^2} = \frac{n+1}{2\sqrt{n^2-1}} $$ As $n \to \infty$, it is clear both sides converge to $\frac12$. By squeezing, we obtain $$\lim_{n\to\infty} \sum_{k=1}^n \arcsin\frac{k}{n^2} = \frac12$$