We known that there exists a constant $C\geq 0$ such that for all $n\geq 1$ : $$|S_n(x)|=\left|\sum _{k=1}^n \frac{\cos (2\pi kx)}{k}\right| \leq C -\log |\sin (\pi x)|, \quad \forall x\in (0,1]. $$ See, for example: Show that $|\sum _{j=1}^n \frac{\cos (2\pi jx)}{j}| \leq C -\log |\sin (\pi x)|$.
Now, if $\lambda_k$ is a sequence of real numbers such that $\lambda_k-\lambda_{k-1}\geq \gamma>0$ is there a similar estimate for $$|S_n(x)|=\left|\sum _{k=1}^n \frac{\cos (2\pi \lambda_k x)}{\lambda_k}\right|?$$
First of all, in the original estimate the interval should be (0, 1 ) instead of (0, 1] because the series in x = 1 isn’t different from x = 0 and diverges.
Anyway, the answer about estimate in (0, 1), of course, is no. We see $\cos(2π λ_k x)$ in the numerator (is it a mistake?), and choosing an arbitrary 0 < q < 1 and $λ_k := k/q$ causes the series to diverge in x = q that belongs to (0, 1). I guess, true original poster’s intent was
$$|S_n(x)|=\left|\sum _{k=1}^n \frac{\cos (2\pi k x)}{λ_k}\right|?$$
for which I don’t know the answer.