Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$

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

$$S_n(x_{n,0}) \nearrow \lim_{n \to \infty}\sum_{k=1}^n \frac{\sin k x_{n,0}}{k} = \lim_{n \to \infty}\frac{\pi}{n+1}\sum_{k=1}^n \frac{\sin \frac{k\pi}{n+1}}{\frac{k\pi}{n+1}} \\ = \int_0^\pi \frac{\sin x}{x} \, dx$$