Yes.

I say the next thing:

\(\displaystyle \Sigma^{\infty}_{n=1} \frac{cos{nx}-cos{(n+1)x}}{n}=2sin(\frac{1}{2}x)\Sigma^{\infty}_{n=1}\frac{sin(\frac{2n+1}{2}x}{n}\)

So, I need to prove that for every \(\displaystyle \epsilon >0\), exist \(\displaystyle N(\epsilon)\), for all \(\displaystyle n\), \(\displaystyle n>N(\epsilon)\), and for all \(\displaystyle p\) natural:

\(\displaystyle |S_{n+p}-S_n|<\epsilon\)

I choose \(\displaystyle p=n\).

\(\displaystyle |S_{2n}-S_n|=|\frac{sin(\frac{2(n+1)+1}{2}x)}{n+1}+...+\frac{sin(\frac{4n+1}{2}x)}{2n}|<|\frac{sin(\frac{2(n+1)+1}{2}x)}{n+1}|+...+|\frac{sin(\frac{4n+1}{2}x)}{2n}|<\frac{1}{n+1}+...+\frac{1}{2n}<n\frac{1}{n+1}<\frac{n}{n}<1\)

My \(\displaystyle \epsilon\) is not containing \(\displaystyle n\)...