Show that $S_n(x)=\sum\limits^{n}_{k=1}\frac{\sin(kx)}{k},$ is uniformly bounded. 
Show that
  $$S_n(x)=\sum^{n}_{k=1}\frac{\sin(kx)}{k},$$
  is uniformly bounded in $\mathbb{R}.$


I have 
$$|S_n(x)|=\Bigg|\sum^{n}_{k=1}\frac{\sin(kx)}{k}\Bigg|\leq\sum^{n}_{k=1}\frac{|\sin(kx)|}{k},$$
but that doesn't help, as it doesn't even imply the sequence is bounded. I also tried rewriting a follows
$$S_n=\int\limits^{x}_{0}\sum^{n}_{k=1}\cos(kt)\,dt,$$
but once again I don't see a way to use this to show anything about the boundedness of $S_n.$

Could you please help me with a hint, only a hint, on how to get started on this problem?
Thank you for your time and appreciate any help or feedback provided.
 A: By periodicity it is enough find a uniform bound for $x \in [0,\pi]$.
We have 
$$\tag{*}\left|\sum_{k=1}^n \frac{\sin kx}{k}\right| = \left|\sum_{k=1}^m \frac{\sin kx}{k}+ \sum_{k=m}^n \frac{\sin kx}{k}\right| \leqslant \sum_{k=1}^m \frac{|\sin kx|}{k}+ \left|\sum_{k=m+1}^n \frac{\sin kx}{k}\right|$$ 
where $m = \lfloor \frac{1}{x}\rfloor$ and $m \leqslant \frac{1}{x} < m+1$.
Since $|\sin kx| \leqslant k|x| = kx$, we have for the first sum on the RHS of (*),
$$ \sum_{k=1}^m \frac{|\sin kx|}{k} \leqslant mx \leqslant 1$$
Hint for remainder of proof
The second sum can be handled using summation by parts and a well-known bound for $\sum_{k=1}^n \sin kx$.
A: For convenience, denote $X\lesssim Y$ or $Y\gtrsim X$ for $X\leq CY$ where $C>0$ is some constant that is independent of $n$ and $x$.
One has actually a stronger result: there exists a constant $C$ such that $|S_n(x)|\leq C$ for all $x\in\mathbb{R}$ and all positive integer $n$.
Since $|S_n(x+\pi)| = |S_n(x)|$ for all $x\in\mathbb{R}$, one can reduce the problem to estimates on the interval $[0,\pi]$. Henceforth, we assume that $x\in[0,\pi]$.
One can control the sum $S_n$ by splitting it into two parts: $kx\leq1$ and $kx>1$, where $k=1,2,\cdots,n$.
For $kx\leq 1$, we have $|\sin(kx)|\leq kx$ and thus
$$
\sum_{1\leq k\leq n\\kx\leq 1}\frac{|\sin(kx)|}{k}\leq
\sum_{1\leq k\leq n\\kx\leq 1}\frac{kx}{k}
\leq
\sum_{1\leq k\leq n\\kx\leq 1}x\leq 1\tag{1}
$$
(Note: up to this point, this is essentially rephrasing an argument in the accepted answer. It may be helpful to look at it at different perspectives.)
Now we handle the part where $kx >1$. Applying the trigonometric identity
$$
2\sin a\sin b = \cos(a-b)-\cos(a+b),
$$
one has
$$
2\sin\frac{x}{2}\sin(kx) = \cos(\frac{x}{2}-kx)-\cos(\frac{x}{2}+kx)
$$ and thus
$$
\begin{align}
2\sin\frac{x}{2}\cdot\sum_{1\leq k\leq n\\kx> 1}\frac{\sin(kx)}{k}
&=\sum_{1\leq k\leq n\\kx> 1}\frac{\cos(\frac{x}{2}-kx)-\cos(\frac{x}{2}+kx)}{k}\\
&=\sum_{1\leq k\leq n\\kx> 1}\frac{\cos((k-\frac12)x)-\cos((k+\frac12)x)}{k}
\end{align}
$$
Note that this is very much like a telescoping sum, except that one has the extra denominator $k$. Now we rearrange the sum as follows. Let $k_0$ be smallest $k$ such that $kx>1$. Then
$$
\begin{align}
&\sum_{1\leq k\leq n\\kx> 1}\frac{\cos((k-\frac12)x)-\cos((k+\frac12)x)}{k}
\\=&\frac{\cos((k_0-\frac12)x)}{k_0}
+\sum_{k=k_0+1}^n\cos((k-\frac12)x)\left(\frac{1}{k}-\frac{1}{k-1}\right)
- \frac{\cos((n+\frac12)x)}{n}\;\tag{2}
\end{align}
$$
Moreover,
$$
\left|\frac{\cos((k_0-\frac12)x)}{k_0}\right|+
\left|\frac{\cos((n+\frac12)x)}{n}\right|\lesssim 1\tag{3}
$$
and
$$
\left|\sum_{k=k_0+1}^n\cos((k-\frac12)x)\left(\frac{1}{k}-\frac{1}{k-1}\right)
\right|\lesssim \sum_{m=1}^\infty\frac{1}{m^2}\tag{4}
$$
Also, for $x>\frac{1}{k_0}$, 
$$
\sin\frac{x}{2}\gtrsim 1\tag{5}
$$
Combining (1)--(5) and using the triangle inequality, one obtains
$$
|S_n(x)|\lesssim 1\;.
$$
A: HINT:
$S_n$ is the partial Fourier series of a "Sawtooth wave".
