Proving that the limit of the following series is $\frac{2}{\pi}$ 
$$\lim_{n \to \infty}(\sin(\frac{\pi}{n})+ \sin(\frac{2\pi}{n}) + \cdots + \sin(\frac{n\pi}{n}))= \frac{2}{\pi}$$

My attempt:
$$\lim_{n \to \infty}(\sin(\frac{\pi}{n})+ \sin(\frac{2\pi}{n}) + \cdots + \sin(\frac{n\pi}{n}))= \lim_{n \to \infty}\frac{\pi}{n}(\frac{\sin(\frac{\pi}{n})}{\frac{\pi}{n}}+ \frac{2\sin(\frac{2\pi}{n})}{\frac{2\pi}{n}} + \cdots + \frac{n\sin(\frac{n\pi}{n})}{\frac{n\pi}{n}}) \\= \lim_{n\to\infty}\frac{\pi}{n}(1+2+\cdots + n) = \pi\sum_{r=1}^n(\frac{r}{n}) = \pi\int_{0}^{1}xdx = \frac{\pi}{2} $$
This has been my attempt. Is this correct?
 A: No, you are not correct: after the second equality you evaluated the limit of the terms separately, but the number of such terms depends on the variable $n$ (you can't do that), moreover it should be
$$\sum_{r=1}^n(\frac{r}{n})=\frac{n(n+1)}{2n}=\frac{n+1}{2}.$$
A more direct approach. Note that as $n\to \infty$,
$$\frac{1}{n}\sum_{k=1}^n\sin(k\pi/n)\to\int_0^{1}\sin(\pi x)\,dx=\frac{2}{\pi}$$
Therefore
$$\lim_{n\to \infty} \sum_{k=1}^n\sin(k\pi/n)=+\infty.$$
A: Your second equality is not correct. In fact with what you have done you will get the limit as $\infty$.
$\frac 1 n (1+2+\cdots+n)=\frac 1  n \frac {n(n+1)} 2 \to \infty$.
A: No, this isn't correct. We know:
$$\lim_{n\to \infty} \frac 1n \sum_{r=1}^n f\left(\frac rn \right)=\int_{0}^1 f(x)\ dx$$
But this is not the form in which you have represented the series. Moreover, there is a simpler alternative approach without involving Riemann sums. Note that the angles in the series form an AP. So simply use the formula:
$$\sum_{r=1}^n \sin(a+(r-1)d)=\frac {\sin \left(a+\frac {(n-1)d}{2}\right) \cdot \sin \frac {nd}{2}}{\sin \frac d2}$$
Now let $a=d=\frac {\pi}{n}$ and evaluate the limit of the closed form.
Aliter:
I shall propose a hint for another alternative:
Let $$S=\sum_{r=1}^{\infty} \sin \frac {r\pi}{n}$$ and $$C=\sum_{r=1}^{\infty} \cos \frac {r\pi}{n}$$
Then, from Euler's formula, $C+iS$ forms an infinite GP which can be easily evaluated. Its imaginary part will be the required answer.
