Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$ I would like to prove that for every $x \in \mathbb{R}, n$ natural, we have $$\!\left|\sin(x)+\frac{\sin(2x)}{2}+\ldots+\frac{\sin{(nx)}}{n}\right| \le \int_{0}^{\pi}\frac{\sin x}{x}\,dx$$
We can of course restrict our attention for $x \in [0, 2\pi]$.
So far I tried a few things: letting $S_n(x)$ denote the quantity inside the absolute value, we have$$S_n'(x) = \sum_{k=1}^{n}\cos(kx) = -\frac1{2}+\frac{\sin(n+\frac12)x}{2\sin\frac{x}{2}}$$
and hence $S_n(x)$ has critical points at $x = \frac{2k\pi}{n}$ or $x = \frac{(2k+1)\pi}{n+1}$ where $k$ is an integer. Moreover:$$S_n(x) = -\frac{x}{2}+\int_{0}^{x}\frac{\sin(n+\frac12)t}{2\sin\frac{t}{2}}\,dt$$
I also know that the function $F(x) = \int_{0}^{x}\frac{\sin(t)}{t}\,dt$ has an absolute maximum for $x = \pi$. I struggle to bound the integral in the expression for $S_n$. Can you help me? Or maybe there's a different approach to solve this?
 A: This is the partial sum of the Fourier series for the sawtooth wave where $S(x) = \sum_{n=1}^\infty \frac{\sin nx}{n} = \frac{\pi-x}{2}$ on $(0,2\pi)$.  The waveform has a discontinuity at $x=0$ and $S(x) \to \frac{\pi}{2}$ as $x \to 0+$.
Note that
$$\int_0^\pi \frac{\sin x}{x} \, dx  = 1.85194\ldots > \frac{\pi}{2}$$
However, the partial sums overshoot the waveform (Gibbs phenomena) and it is proved here that
$$\limsup_{n \to \infty}\sup_{x \in [0,\pi]} S_n(x) \geqslant  \int_0^\pi \frac{\sin x}{x} \, dx $$

We can prove that $\displaystyle|S_n(x)| \leqslant \int_0^\pi \frac{\sin x}{x} \, dx$ for all $n \in \mathbb{N}$ and $x \in [0,2\pi]$.
By periodicity and anti-symmetry it is enough to consider the interval $[0,\pi]$. Here the relative maxima of $S_n (x)$ occur at $x_{n,m} = \frac{2m+1}{n+1}\pi$ for $m = 0,1, \ldots, \lfloor\frac{n-1}{2} \rfloor$.
It can be shown that $S_n(x_{n,m})  > S_n(x_{n,m+1})$ so that there is an absolute maximum of $S_n(x)$ on $[0,\pi]$ at $x_{n,0} = \frac{\pi}{n+1}$. It can also be shown that $S_n(x_{n,0})$ is an increasing sequence, such that
$$S_n(x_{n,0}) \nearrow \lim_{n \to \infty}\sum_{k=1}^n \frac{\sin k x_{n,0}}{k} = \lim_{n \to \infty}\frac{\pi}{n+1}\sum_{k=1}^n \frac{\sin \frac{k\pi}{n+1}}{\frac{k\pi}{n+1}} \\ = \int_0^\pi \frac{\sin x}{x} \, dx$$
