What does this sum tend to approach to?

Question

philomath · Answer

$$\Large \frac{\pi}{1000}\sum_{j=1}^{1000}\sin\left(\frac{\pi \times j}{1000}\right)$$

philomath · Answer

The answer is 2 but I want to know how to get there

zarkon · Answer

$$\int_{0}^{\pi}\sin(x)dx=2$$

philomath · Answer

@Zarkon How did you know that it was just the integral? I know of a rule that for very small n, sin(pi/n) = pi/n

philomath · Answer

Unless that is the wrong rule and it was written some other way

zarkon · Answer

$$\lim_{n\to\infty}\sum_{j=1}^{n}f(x_j)\Delta x=\int\limits_{a}^{b}f(x)dx$$

so for large $n$ we have 

$$\sum_{j=1}^{n}f(x_j)\Delta x\approx\int\limits_{a}^{b}f(x)dx$$

let $f(x)=\sin(x)$ and $\Delta x=\frac{b-a}{n}$

let $a=0, b=\pi $ and $n=1000$

philomath · Answer

First principles ahhhh. And you can just say the pi*j/1000 is equal to sin(x) or in this case sin(j)?

zarkon · Answer

using the notation above with $x_j=a+j\Delta x$
we get 

$$\sum_{j=1}^{n}f(x_j)\Delta x=\sum_{j=1}^{n}f\left(0+j\frac{\pi-0}{1000}\right)\frac{\pi-0}{1000}$$

$$=\sum_{j=1}^{n}f\left(\frac{j\cdot\pi}{1000}\right)\frac{\pi}{1000}=\sum_{j=1}^{n}\sin\left(\frac{j\cdot\pi}{1000}\right)\frac{\pi}{1000}$$

which is what you have above.

philomath · Answer

Awesome! Thank you again