Please check my answer to $\sum\limits_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$ $$\sum_{i=1}^n \frac{\sin{(ix)}}{i} &lt 2\sqrt{\pi}$$
I have this answer, please let me know if there is a more beautiful proof. 
My answer: 
at first, we prove two inequalities: 


*

*If $x\in (x,\pi)$ then $\sin x \leq x$ 

*When $x\in(0,\frac{\pi}{2})$, $\sin x \geq \frac{2x}{\pi}$


1) first, let $y = \sin x -x $ 
$y^{\prime} = \cos x -1 \leq 0$
so $\sin x - x \leq \sin 0 -0 = 0$
which can be rewritten as 
$\sin x \leq x$
2) Let $y=\sin x - \frac{2x}{\pi}$ 
thus $y^{\prime} = \cos x - \frac{2}{\pi}$ because $x\in (0, \frac{\pi}{2})$
so y at first decreases and then increases on the boundary of $x \in (0,\frac{\pi}{2})$  
so $ \sin x - \frac{2}{\pi}\leq \max \{{\sin 0 - \frac{2}{\pi}0, \sin (\frac{\pi}{2} - \frac{2}{\pi}\frac{\pi}{2}) \}}$ 
so $\sin x \leq \frac{2x}{\pi}$
Then select $M\in N$ 
$\frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m+1} + \ldots + \frac{\sin ((m+n)x)}{m+n} \leq  \frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m} + \ldots + \frac{\sin ((m+n)x)}{m} $ 
=> $\frac{1}{2M} \times \frac{\sin ((m-\frac{1}{2})x) - \sin ((n+\frac{1}{2})x)}{\sin \frac{x}{2}} &lt \frac{1}{m \times \sin \frac{x}{2}} \times \sin x + \frac{\sin 2x}{2} + \ldots + \frac{\sin ((m-1)x)}{m-1} &lt x + \frac{2x}{2} + \ldots + \frac{(m-1)x}{m-1} $
so just need to prove that 
$(m-1)x + \frac{1}{m \times \sin \frac{x}{2}} \leq 2\sqrt{\pi}$ 
select M which satisfies 
$ \frac{\sqrt{\pi}}{x} \leq m &lt \frac{\sqrt{\pi}}{x} + 1 $ 
so $ (m-1)x &lt [ \frac{\sqrt{\pi}}{x} \times x = \sqrt{\pi} ] $
thus 
$\frac {1}{m \times \sin(\frac{x}{2})}\leq[ \frac{1}{\sqrt{\pi}}\times \frac{2}{\frac{ \sin (0.5x)}{0.5x}} = \frac{1}{\sqrt{\pi}} \times \frac{2}{\frac{\sin 0.5x}{0.5x}} ]$ 
because $x\in (0, \pi)$ thus $\frac{x}{2} \in (0, \frac{\pi}{2})$ 
$ (m-1)x + \frac{1}{m \times \sin(0.5x)} \leq 2\sqrt{\pi} $
thanks, for viewing and commenting. 
ps. I'm still learning latex and mathematics, so my answer isn't pretty to read, nor is the latex I wrote. 
 A: The best it is possible to state is:
$$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots$$
Call $f_N(x)=\sum_{n=1}^N \frac{\sin(nx)}{n}$: it is a $2\pi$-periodic function converging to $\frac{\pi-x}{2}$ in $L_2\left([0,2\pi]\right)$. 
Since:
$$\frac{d f_N(x)}{dx}= \frac{\cos\left(\frac{N+1}{2}x\right)\sin\left(\frac{N}{2}x\right)}{\sin\left(\frac{x}{2}\right)},$$
we know that $f_N(x)$ has $2n$ stationary points in $[0,2\pi]$, local maxima in $x=\frac{(2k+1)\pi}{N+1}$, the first one occurring in $x_N=\frac{\pi}{N+1}$. Once we prove that the value of $f_N(x)$ in any other local maximum is less than $f_N(x_N)$, and that $f_N(x_N)$ is an increasing sequence (I still must find a convincing proof of this two facts, but they look not too hard to deal with and strongly supported by computer inspection) the best bound we can hope in is:
$$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\lim_{N\to +\infty}\sum_{n=1}^{N}\frac{\sin\left(\frac{\pi x}{N+1}\right)}{n},$$
where the RHS a Riemann sum associated with:
$$\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots<\frac{13}{7},$$
QED.
A: I was correct. 
Read the comments for better ideas. 
Answering this out of a need for my question to have an answer, and I wrote a correct answer in my post. 
Yea, for me. 
