Calculate the limit $\lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}\sin^2\frac{r}{n}$

Question

How do I calculate the limit $$\lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}\sin^2\frac{r}{n}$$ It looks like a Riemann sum without the limit, but I don't have a really good understanding of Riemann sum so I'm stuck.

I guess $\Delta x = \frac{1}{n}$, and $f(x)=\sin^2\frac{x}{n}$. Then maybe I have to use (from my textbook)

$$Riemann\ Sum =\frac{b-a}{n} \sum_{k=1}^n f\left(a + (b-a)\frac{k}{n}\right)$$ to obtain $a$ and $b$, which would be used as $\int_{a}^{b}f(x)$?

So $b-a = 1$, and $x=a + (b-a)\frac{k}{n}$? How do I do this?

Answer 1

I like to look at it this way:

Take the lower limit of the sum:

$\text{Lower limit}\stackrel{(r=1,n\to\infty)}=\frac1n=0$

Take the upper limit of the sum:

$\text{Upper limit}\stackrel{(r=n,n\to\infty)}=\frac nn=1$

Thus, it becomes:

$$\lim_{n\to\infty}\frac1n\sum_{r=1}^n\sin^2(r/n)=\int_0^1\sin^2(x)\ dx$$

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Following your guidelines, you could note that we should have $f=\sin^2$, and thus,

$$a=0$$

$$b-a=1$$

Since the summand is of the form

$$\sin^2\left(0+(1)\frac rn\right)$$