$\lim_{n\to \infty}\frac{S_1+S_2+S_3+.....+S_n}{n}=\frac{1}{2}\cot\frac{\theta}{2}$ Let $S_n=\sin \theta+\sin 2\theta+\sin 3\theta+.......+\sin n\theta$.Prove that
$\lim_{n\to \infty}\frac{S_1+S_2+S_3+.....+S_n}{n}=\frac{1}{2}\cot\frac{\theta}{2}$

$\lim_{n\to \infty}\frac{S_1+S_2+S_3+.....+S_n}{n}=\lim_{n\to \infty}\frac{\sin \theta+(\sin \theta+\sin 2\theta)+(\sin \theta+\sin 2\theta+\sin 3\theta)+.....+(\sin \theta+\sin 2\theta+\sin 3\theta+.......+\sin n\theta)}{n}$
$=\lim_{n\to \infty}\frac{n\sin \theta+(n-1)\sin 2\theta+(n-2)\sin  3\theta+.....+2\sin (n-1)\theta+\sin n\theta}{n}=\lim_{n\to \infty}\frac{\sum_{k=1}^{n}(n-k+1)\sin k\theta}{n}$
and then i stuck.Please help me get through.
 A: \begin{align}
S_{n,n}&=S_1+S_2+...+S_n\\
&=\sum_{j=1}^n\sum_{k=1}^j\sin(k \theta)\\
&=\sum_{k=1}^n\sum_{j=k}^n\sin(k \theta)\\
&=\sum_{k=1}^n(n+k-1)\sin(k \theta)\\
&=\Im\Big(\sum_{k=1}^n(n+k-1)e^{ik \theta}\Big)\\
&=(n-1)\Im\Big(\sum_{k=1}^ne^{ik \theta}\Big)+\Im\Big(\sum_{k=1}^nke^{ik \theta}\Big)\\
&=(n-1)\Im\Big(\frac{e^{i\theta}-e^{i(n+1)\theta}}{1-e^{i\theta}}\Big)+\Im\Big(\frac{(n e^{i\theta}-n-1) e^{i(n+1)\theta}+e^{i\theta}}{(1-e^{i\theta})^2}\Big)\\
&=(n-1)\Im\Big(\frac{\cos(\theta)+i\sin(\theta)-\cos((n+1)\theta)-i\sin((n+1)\theta)}{1-\cos(\theta)-i\sin(\theta)}\Big)+\Im\Big(\frac{(n (\cos(\theta)+i\sin(\theta))-n-1) (\cos((n+1)\theta)+i\sin((n+1)\theta))+\cos(\theta)+i\sin(\theta)}{(1-\cos(\theta)-i\sin(\theta))^2}\Big)\\
\end{align}
now using $\Im\frac{a+ib}{c+id}=\frac{bc-ad}{c^2+d^2}$ everywhere we obtain 
$$\frac1nS_{n,n}=\frac{4 \sin ^2\left(\frac{\theta }{2}\right) ((n-1) \sin (\theta )+2 n \sin (n \theta )+(1-2 n) \sin ((n+1) \theta ))}{n\Big(1-\cos (\theta)\Big)^2}$$
... and we are happy :-)
A: You wrote in a comment that the formula
$$
S_n=\sum_{k=1}^n \sin k\theta=\frac{\sin (n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}
$$
was known. Using, the addition formula for sine, and the formulas $\sin^2 t=(1-\cos 2t)/2$ and $\cos t\sin t=\frac{1}{2}\sin 2t$, we have
$$
\begin{aligned}
S_n&=\sin(n\theta/2)^2\frac{\cos(\theta/2)}{\sin(\theta/2)}+\sin(n\theta/2)\cos(n\theta/2)\\
&=\frac{1}{2}\cot(\theta/2)-\frac{1}{2}\cot(\theta/2)\cos(n\theta)+\frac{1}{2}\sin(n\theta).
\end{aligned}
$$
Since you knew the formula for the sums of sines, you probably do for the sums of cosines,
$$
\sum_{k=1}^n \cos k\theta=\frac{\sin(n\theta/2)\cos((n+1)\theta/2)}{\sin(\theta/2)}.
$$
Thus,
$$
\frac{1}{n}\sum_{k=1}^n S_k=\frac{1}{2}\cot(\theta/2)-\frac{1}{2}\cot(\theta/2)\frac{\sin(n\theta/2)\cos((n+1)\theta/2)}{n\sin(\theta/2)}+\frac{1}{2}\frac{\sin (n\theta/2)\sin((n+1)\theta/2)}{n\sin(\theta/2)}.
$$
I'm sure you can conclude from here. Also, I leave it to you to check that nothing strange happens when $\sin(\theta/2)=0$.
