Show that $|\sum _{j=1}^n \frac{\cos (2\pi jx)}{j}| \leq C -\log |\sin (\pi x)| $ Show that there exists a constant $C\geq 0$ such that for all $n\geq 1$ : $$|S_n(x)|=|\sum _{j=1}^n \frac{\cos (2\pi jx)}{j}| \leq C -\log |\sin (\pi x)|, \quad \forall x\in ]0,1]. $$
Since $S_n(x)$ converge uniformly on each interval $[\epsilon , 1]$ with $\epsilon >0$, it is suffisant to show that the result is true on a neighbourhood of $0$.
I tried to use an Abel transformation and use the fact that $|\sum _{j=1}^n\cos (2\pi jx)|\leq C'|\sin (\pi x)^{-1}|$. But this majoration is too crude and it doesn't work.
 A: This is indeed an exercise about Fourier series. We have that:

$$\log\sin(\pi x) = -\log 2-\sum_{n\geq 1}\frac{\cos(2\pi n x)}{n} $$

holds as an identity over $L^2((0,1))$, since:
$$ \int_{0}^{1}\log\sin(\pi x)\,dx = -\log 2$$
follows from considering Riemann sums and the identity:
$$ \prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n},$$
while over $|z|<1$ we have:
$$\sum_{n\geq 1}\frac{z^n}{n} = -\log(1-z) $$
so we can evaluate the previous identity in $z=e^{2\pi i x}$ by Abel's lemma, then take the real part.
A: I think I've found an alternative proof.
We can write $\sum _{j\geq 1}\frac{\cos (2\pi jx)}{j} =\sum _{j\leq \lambda} \frac{\cos (2\pi jx)-1}{j}+ \sum _{j\leq \lambda}j^{-1}+\sum _{j>\lambda } \frac {cos(2\pi jx)}{j}=A_1+A_2+A_3.$
Since $|\cos (x) -1|\leq x^2/2$ we have : $$ |A_1|\leq \sum _{j\leq \lambda}4\pi ^2 jx^2 \leq C_1\lambda ^2x^2 $$
and : $$ |A_2| \leq  \log (\lambda ) +1$$
By using Abel transform we can show that : $$ |A_3| \leq C_3 (\lambda x)^{-1}.$$
By taking $\lambda =x^{-1}$, we obtain that : $\sum _{j\geq 1}\frac{\cos (2\pi jx)}{j} \leq C' +\log (1/x) \leq C -\log (| \sin (\pi x)|).$  
Do you think my proof is correct ?
