$\lim\limits_{n\to\infty}\frac{\sum\limits_{i=1}^{n}\sin(\frac{\pi i}{n})}{n}$ I can solve it by thinking about it as the definition for the definite integral 
$\int_0^1\sin(\pi x)dx$ . Are there any other ways to solve it?
 A: Hint
With $j=\sqrt{-1}$:
$$\Large\sin{\pi i\over n}={1\over 2j}\left(e^{j\pi i\over n}-e^{-j\pi i\over n}\right)$$and use the geometric sum.
A: Hint:
You can use the trigonometric formula for arcs in arithmetic progression:
$$ \sin \theta + \sin 2 \theta + \dots + \sin n \theta  = \frac{\sin \cfrac{(n + 1)\theta }{2}}{\sin \cfrac{\theta }{2}}\,\sin \frac{n \theta }{2},$$
set $\theta=\dfrac\pi n$ and do some (high-school) trigonometry.
You only need to know the standard limit:
$$\lim_{x\to 0}\frac{\sin(ax)}x=a.$$
A: Keeping in mind Euler identity, consider
$$S_n=\sum_{k=1}^{n}e^{i\frac{\pi k}{n}}=\sum_{k=1}^{n}\left(e^{\frac{i\pi }{n}}\right)^k=\frac{2 e^{\frac{i \pi }{n}}}{1-e^{\frac{i \pi }{n}}}$$ Expand the complex numbers, simplify to get
$$S_n=-1+i \cot \left(\frac{\pi }{2 n}\right)\implies \sum_{k=1}^{n}\sin\left(\frac{\pi k}{n}\right)=\cot \left(\frac{\pi }{2 n}\right)$$
$$T_n=\frac{\sum_{k=1}^{n}\sin\left(\frac{\pi k}{n}\right) } n=\frac 1n\cot \left(\frac{\pi }{2 n}\right)$$ Now, using Taylor expansion
$$T_n=\frac{2}{\pi }-\frac{\pi }{6 n^2}-\frac{\pi ^3}{360
   n^4}+O\left(\frac{1}{n^6}\right)$$ which shows the limit and how it is approached.
Moreover, this gives a shortcut method for computing $T_n$. Try for $n=6$ (far away from $\infty$).
The exact result
$$T_6=\frac{2+\sqrt{3}}{6} \approx 0.62200847$$ while the above truncated expansion gives
$$T_6 \sim \frac{2}{\pi }-\frac{\pi }{216}-\frac{\pi ^3}{466560}\approx 0.62200890$$
