Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$. Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.
It seems to me like a Riemann integral definition:
$\sum_{k=1}^n\frac{\sin{\frac{k\pi}{n}}}{n}=\frac1n(\sin{\frac{\pi}n}+...+\sin\pi)$
So $\Delta x=\frac 1n$ and $f(x)=\sin(\frac{\pi x}n)$ (Not sure about $f(x))$
How do i proceed from this point?

I think it should be 
$$
\int_{0}^{1}\sin \pi xdx=\frac{2}{\pi}
$$
EDIT OK, the point here is the direct implementation of Riemanns sums with $d x = \frac{b-a}{n}, \ b=1, \ a=0$ and $\frac{k}{n}=x$
