Does the expression $\lim_{n\to\infty} (1/n)(\sin(\pi/n)+\sin(2\pi/n)+\cdots+\sin(n\pi/n))$ converge or diverge? I am having a tough time deciding which Convergence test should I use to determine if the expression converges or diverges?
the expression is:
$$\lim_{n\to\infty}\frac{1}{n}\left(\sin\frac{\pi}{n}+\sin\frac{2\pi}{n}+\cdots+\sin{n\pi}{n}\right)$$
I tried coding it in Mathematica and it seems like the series converges, but I cannot decide which Convergence test should I use to determine that? Any help is appreciated.
 A: Hint. If $\sin (x/2)\ne 0$ then $$\sum_{j=1}^n\sin jx=(\sin (x/2))^{-1}\sum_{j=1}^n(\sin jx)(\sin(x/2))=$$ $$=(\sin (x/2))^{-1}\sum_{j=1}^n(1/2)(\;\cos(jx-x/2)-\cos (jx+x/2)\;)=$$ $$=(\sin (x/2))^{-1}(1/2)(\;\cos (x/2)-\cos (nx+x/2)\,).$$ An example of a telescoping series, i.e. $\sum_{j=1}^n(\;F(j)-F(j+1)\;)=F(1)-F(n+1).$
We can also write $\sum_{j=1}^n\sin jx=$ $(2i)^{-1}\sum_{j=1}^n\exp (ijx)-(2i)^{-1}\sum_{j=1}^n\exp (-ijx)$ (where $i^2=-1$) and sum these two geometric series, and after a little re-arrangement, get the result above.
A: If we've developed the theory of integration and have proven that $\sin$ is integrable over $[0,\pi]$, then the existence of the limit can be proven by writing
\begin{align*}
\frac{1}{n}\left(\sin\left(\frac{\pi}{n}\right)+\sin\left(\frac{2\pi}{n}\right)+\sin\left(\frac{3\pi}{n}\right)+\cdots+\sin\left(\frac{n\pi}{n}\right)\right) &=\frac{1}{n}\sum_{i=1}^{n}\sin\left(\frac{i\pi}{n}\right)\\
&= \frac{1}{\pi}\sum_{i=1}^{n}\sin\left(0+\frac{i\pi}{n}\right)\frac{\pi}{n}\\
&= \frac{1}{\pi}\sum_{i=1}^{n}\sin\left(0+i\frac{\pi-0}{n}\right)\frac{\pi-0}{n}
\end{align*}
and observing that the sum $\sum_{i=1}^{n}\sin\left(0+i\frac{\pi-0}{n}\right)\frac{\pi-0}{n}$ is simply a Riemann sum for $\sin$ over $[0,\pi]$, where the subintervals $[x_{i-1},x_i]$ are of equal length and the tags $x_{i}^{*}$ are chosen to be the right-hand endpoints of each subinterval.
Letting $n$ tend to infinity and exploiting the integrability of $\sin$ gives
\begin{align*}
\lim_{n\to\infty}\frac{1}{n}\left(\sin\left(\frac{\pi}{n}\right)+\sin\left(\frac{2\pi}{n}\right)+\sin\left(\frac{3\pi}{n}\right)+\cdots+\sin\left(\frac{n\pi}{n}\right)\right) &= \lim_{n\to\infty}\frac{1}{\pi}\sum_{i=1}^{n}\sin\left(0+i\frac{\pi-0}{n}\right)\frac{\pi-0}{n}\\
&=\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\text{ }dx
\end{align*}
If we've also developed the theory of differential calculus and established that the derivative of $\cos$ is $-\sin$, we can take this a step further and evaluate the limit in closed-form using the Fundamental Theorem of Calculus.
$$\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\text{ }dx = \frac{1}{\pi}\left(-\cos(\pi)-(-\cos 0)\right)=\frac{1-(-1)}{\pi}=\frac{2}{\pi}$$
A: Geometrically, it is clear that the sum is going to converge to the barycenter of a semi-circumference.
A: Because of the symmetry of $sin(x)$ around $x=\frac{\pi}{2}$ the series can be split $=\frac{2}{n}(\sum\limits_{k=1}^{\frac{n}{2}}sin(\frac{k\pi}{n}))\lt \frac{2\pi}{n^2}\sum\limits_{k=1}^{\frac{n}{2}}k\to \frac{\pi}{4}$
