the series $\sum_{1}^{\infty} \frac{\cos(nx)}{\{\log(n+1)\}^x}$ is uniformly convergent on any closed interval $[a,b]$ lying within $(0,2\pi)$.

To show that the series $\sum_{1}^{\infty} \frac{\cos(nx)}{\{\log(n+1)\}^x}$ is uniformly convergent on any closed interval $[a,b]$ lying within $(0,2\pi)$.

My Try: Let us consider $u_n(x) =\cos(nx), x \in [a,b] \subset (0,2\pi);$ and $v_n(x) = \frac{1}{\{\log(n+1)\}^x}$.

Then $v_n(x)$ is a monotonic decreasin g sequence converging to $0$.

Let $s_n(x) = u_1(x) + u_2(x)+...+u_n(x) = \frac{sin (nx/2) \cos \frac{(n+1)x}{2}}{sin(x/2)}$. For each $n \in \mathbb N, |s_n(x)| \leq |\frac{1}{sin \frac{x}{2}}|$.

Since $\sin \frac{x}{2} \neq 0$.....(details can be done!!)

Thus $s_n(x)$ is uniformly bounded on $[a,b]$. By Dirichlet Test the series $\sum_{1}^{\infty} \frac{\cos(nx)}{\{\log(n+1)\}^x}$ is uniformly convergent on $[a,b] \subset (0,2\pi)$.

Is the proof correct?