$|\sum_{k=1}^n \frac{\sin(kx)}{k}| \leq \frac{\pi}{2}+1$ I want to prove that $$\left|\sum_{k=1}^n \frac{\sin(kx)}{k}\right| \leq \frac{\pi}{2}+1.$$ for each $n\in\mathbb N$ and $x\in (0,2\pi)$.
I know that the sum is inside is the partial sum of the fourier series of the periodic extension of $f: [0,2\pi)\to\mathbb R, x\mapsto \frac{x-\pi}{2}$ with $f(0)=0$.
Sadly, I can't use the proof of the Gibbs phenomenon. Is there any other way?
 A: Just a sketch of a possible approach, quite close to one of the usual proofs of the Fejer-Jackson inequality. By periodicity and symmetry, we just need to prove the inequality over $(0,\pi)$.
We have:
$$\sum_{n=1}^{N}\frac{\sin(nx)}{n}=\int_{0}^{x}\sum_{n=1}^{N}\cos(nt)\,dt=-\frac{x}{2}+\int_{0}^{x/2}\frac{\sin((2N+1)u)}{\sin(u)}\,du \tag{1}$$
and we may explicitly locate the stationary points of the LHS, the first one appearing at $x=\frac{2\pi}{2N+1}$.
The RHS of $(1)$ evaluated in such a point is, by the concavity of the sine function in a right neighbourhood of the origin:
$$-\frac{\pi}{2N+1}+\int_{0}^{\frac{\pi}{2N+1}}\frac{\sin((2N+1)u)}{\sin u}\,du\leq\frac{\pi}{(2N+1)\sin\frac{\pi}{2N+1}}\int_{0}^{\pi}\frac{\sin u}{u}\,du \tag{2}$$
and the RHS of $(2)$ is less than $2.24<\frac{\pi}{2}+1$ for any $N\geq 1$. So we just need to prove that the first stationary point is indeed an absolute maximum. That is quite natural, since the LHS of $(1)$ is a truncated Fourier series for a smooth decreasing function over $(0,\pi)$. To fill the gaps is left to you.
