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?
 A: HINT:
Using this, $$\sum_{1\le r\le n}\sin \frac{\pi r}n=\frac{\sin \left( n\cdot \frac{\pi}{2n}\right)}{\sin \frac \pi{2n}}\cdot \sin \left(\frac{2\frac\pi n+(n-1)\frac{\pi}n}2\right)=\frac{\sin\left(\frac{(n+1)\pi}{2n}\right)}{\sin \frac \pi{2n}}$$
Now, $\sin\left(\frac{(n+1)\pi}{2n}\right)=\sin\left(\frac\pi2+\frac\pi{2n}\right)=\cos \frac\pi{2n}\implies \lim_{n\to\infty}\sin\left(\frac{(n+1)\pi}{2n}\right)=1$
$$\lim_{n\to\infty}\frac 1{n\sin \frac \pi{2n}}=\lim_{h\to0}\frac {2h}{\pi\sin h }(\text{ putting } \frac \pi{2n}=h)$$
$$\implies \lim_{n\to\infty}\frac 1{n\sin \frac \pi{2n}}=\frac2\pi\cdot\lim_{h\to0}\frac h{\sin h }=\frac2\pi$$
A: Ideas: take the partition $\,P_n\,$ of the unit interval
$$P_n:=\left\{0=x_0\,,\,x_1=\frac1n\,,\ldots,\,x_k=\frac kn\,,\ldots,\,x_n=\frac nn=1\right\}$$
and choose points
$$c_i:=\frac in\;,\;\;1\le i\le n$$
Since we already know  that $\,\sin \pi x\,$ is continuous everywhere and Riemann integrable in any closed, finite interval, we can apply the Riemann integral definition for the particular partitions and particular points $\,c_i\,$ as above in each subinterval, and get:
$$\lim_{n\to\infty}\frac 1n\sum_{k=1}^n \sin\frac{\pi k}n=\int\limits_0^1 \sin\pi x\,dx=\ldots\ldots$$
A: This is a problem that requires proper handling of the concept of Riemann sum and its relationship to definite integral. Consider the following partition of the interval $[0,1]$:
$$
\mathscr{P}\Big( [0,1]\Big)=\left \{ 0=x_0<x_1=\frac{1}{n}<\ldots < x_k=k\cdot \frac{1}{n}<\ldots<x_{n}=n\cdot\frac{1}{n} \right\}
$$
chose $x^*_k\in [x_{k-1},x_{k}]$ equal to $k\cdot \frac{1}{n}$ and set $\Delta x_k=x_k-x_{k-1}$ .
Then we have
$$
\sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k
\leq
 \sum_{k=1}^{n}\sin( x_k^*\pi)\cdot\Delta x_k
\leq
\sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k
$$
that is
$$
\sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k
\leq
\sum_{k=1}^n\sin{\left(\frac{k\pi}{n}\right )}\frac{1}{n}
\leq
\sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi \pi)\cdot\Delta x_k.
$$
If we have
$$
\lim_{n \to \infty}\sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi \pi)\cdot\Delta x_k
=
 \int_0^1\sin( x\pi)dx
=
\lim_{n \to \infty}\sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k
$$
then 
\begin{align}
\lim_{n\to+\infty}\sum_{k=1}^n\sin{\left(\frac{k\pi}{n}\right )}\frac{1}{n}=
 \int_0^1\sin( x\pi)dx
\end{align}
A: 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 $\frac{k}{n}=x$
A: Remember that if $f: [a,b] \to \mathbb{R}$ is some function we can define the Riemann Sum of $f$ in the interval by choosing a partition $P = \{t_0, \dots, t_n\}$ of $[a,b]$ in a very specific manner: we simply divide the interval $[a,b]$ equally calculating it's length $b-a$ and then dividing by the number of subintervals that we want. So a partition with $n$ subintervals will have $t_i = a + i\Delta t$ where $\Delta t = (b-a)/n$ is the size of each interval. In that case the integral of $f$ on the interval $[a,b]$ will be the limit:
$$\int_a^b f(t)dt=\lim_{n\to \infty} \sum_{i=1}^n f(t_i)\Delta t$$
Look that in your case we are working on the interval $[0,1]$ so for a partition of $n$ subintervals we'll have $\Delta t = 1/n$. Now we want to compute the limit of the following sum:
$$\sum_{i=1}^n \frac{1}{n}\sin{\left(\frac{i\pi}{n}\right)}$$
Note that with this we have:
$$\sum_{i=1}^n \Delta t\sin{\left(i\Delta t \pi\right)}$$
However, since the interval starts on $0$ and since $t_i = a + i\Delta t$, we have that $t_i = i\Delta t$, so we have:
$$\sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t$$
Now on the limit this will become obviously the following integral:
$$\lim_{n\to \infty} \sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t=\int_0^1 \sin(\pi t)dt$$
This integral is computed trivially by setting $u=\pi t$, $dt=du/\pi$ and so $u$ ranges from $0$ to $\pi$ and we have:
$$\lim_{n\to \infty} \sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t=\frac{1}{\pi}\int_0^\pi \sin(u)du=\frac{2}{\pi}$$
