proof verification of $\frac{1}{n}\sum \sin nx$ having a limit I have to prove that $b_n=\frac{1}{n}\sum\limits_{i=1}^n \sin ix$ has a limit.
I'm using the result of an already solved problem which implies the following:
$$\sum_{i=1}^n \sin ix=\frac{\sin{\frac{n+1}{2}x}\sin{\frac{n}{2}x}}{\sin{\frac{x}{2}}}$$
Having that we see that $\sum\limits_{i=1}^n \sin ix$ is a bounded sequence since $\sin{x}$ is bounded by $-1$ and $1$.
Using the property that multiplying a bounded sequence by a sequence converging to $0$ we get a sequence converging to $0$, i.e.
$\lim\limits_{n\to\infty}\frac{1}{n} \sum\limits_{i=1}^n \sin ix=0$.
Is my reasoning correct and if not how do I prove this?
 A: One needs to be careful here since we are dividing by $\sin(x/2)$ and that can be $0$ for $x\in2\pi\mathbb{Z}$. If $x\not\in2\pi\mathbb{Z}$, then $\sin(x/2)\ne0$ and your argument works fine.
However, if $x\in2\pi\mathbb{Z}$, then $\sin(ix)=0$ for all $i\in\mathbb{Z}$, and so the sum is $0$. This case needs to be handled.
Therefore, we get a pointwise limit, but we have not shown whether this limit is uniform or not.
A: View the sum as the imaginary part of
$$\sum_{k=1}^{n}e^{ikx} = e^{ix}\frac{e^{inx} - 1}{e^{ix} -1}.$$
That's bounded if $x\in (0,2\pi).$ If $x=0,$ the imaginary part of the sum is simply $0.$
A: Yes, you did it very well. In a formal way, one would write it perhaps as follows: Since we showed that $\sum_{i=1}^n sin(ix)$ can be written by the above formula, it is  $-1 \leq \sum_{i=1}^n sin(ix) \leq 1$, so also
$$ \frac{-1}{n} \leq b_n \leq \frac{1}{n}.$$ 
Since both the sequences $\frac{-1}{n}$ and $\frac{1}{n}$ converge to $0$ for $n \to \infty$, by the "Sandwich criterion" (or however you want to call this criterion with upper and lower bounds), also
$$ \lim_{n \to \infty} b_n = 0. $$

EDIT: After looking at the other answer, I just realized that $\sum_{i=1}^n \sin(ix)$ is not bounded by $1$! The denominator in your expression can be extremely small, making the fraction as large as you like. But as long as it does not get zero, the argument can be still performed since you still get a finite bound, and $x$ is fixed throughout the argument.
