How to evaluate $\lim\limits_{n\to\infty} {\sin({b\over n})+\sin({2b\over n}) + \ldots + \sin({nb\over n})\over n}$ by relating it to a Riemann sum?

Question

Evaluate the following limit by relating it to a Riemann sum:

$$\lim_{n\to\infty} \frac{\sin\left(\frac{b}{n}\right)+\sin\left(\frac{2b}{n}\right) + \ldots + \sin\left(\frac{nb}{n}\right)}{n}$$

Answer 1

Hint : Set your limit to be L. Take the partition of the interval $[0,b]$ as follows $x_i=ib/n$. So, we have $$L=\frac{1}{b}\int _{0}^b \sin x dx$$

Answer 2

Hint. You may try to identify the different elements of the general form $$\lim_{n\to \infty}\sum_{i=1}^{n}\frac{b-a}{n}\cdot f\left(a+i\frac{b-a}{n} \right)=\int_a^bf(x)dx.$$ Try to answer the questions, what are $$ a=? \qquad b=? \qquad f(x)=? $$

Answer 3

It looks like the values that $\sin(x)$ takes are discrete values from $0$ to $b$ and the argument is incremented by ${b \over n}$, this will be $\Delta x$.

$$\Delta x=\frac{b}{n}$$

Your sum looks like

$$\frac{1}{b}\sum_{k=1}^{n}\sin\left({bk\over n}\right)\Delta x$$

Note that I've multiplied and divided by $b$.

Now passing to the limit

$$\frac{1}{b}\int_{0}^{b}\sin(x)\,{\rm d}x$$