I want to determine if following series is uniformly convergent and on what interval if it is:
$$\sum_{k=2}^\infty \frac{\sin(kx)}{k \ln(k)}$$
I see that $ \frac{|\sin(kx)|}{k \ln(k)} \leq \frac{1}{k \ln(k)}$but can’t use the Weierstrass M-test because $\sum \frac{1}{k \ln(k)}$ diverges.
I know from other answers that $\sum \frac{\sin(kx)}{k}$ converges uniformly on compact intervals $[a,b] \subset (0,2\pi)$ but not on $(0,2\pi)$ itself, but this does not seem to help me.
Best Answer
Unlike $S_n(x) = \sum_{k=1}^n \sin kx$, the sum $\sum_{k=1}^n \frac{\sin kx}{k}$ is uniformly bounded for all $n$ and all $x \in \mathbb{R}$.
Hence, this series converges uniformly for all $x \in \mathbb{R}$ by the Dirichlet test -- since $(\ln k)^{-1} $ converges to $0$ monotonically and uniformly with respect to $x$.
Proving that $\sum_{k=1}^n \frac{\sin kx}{k}$ is uniformly bounded requires some effort. Because of periodicity, we can consider WLOG $x \in (0,\pi)$.
With $m = \lfloor1/x \rfloor$ we have
$$\left|\sum_{k=1}^n \frac{\sin kx}{k}\right| \leqslant \sum_{k=1}^{m}\frac{|\sin kx|}{k} + \left|\sum_{k=m+1}^{n}\frac{\sin kx}{k}\right| $$
For the first sum on the RHS,
$$\sum_{k=1}^{m}\frac{|\sin kx|}{k} \leqslant \sum_{k=1}^{m}\frac{k|x|}{k} = mx < 1$$
The second sum can be bounded as well using summation by parts.
Noting that $|S_n(x)| \leqslant \frac{1}{|\sin(x/2)|}$ and $|\sin(x/2)| \geqslant \frac{2}{\pi}\frac{x}{2} = \frac{x}{\pi} $ for $x \in (0,\pi)$ we have,
$$\left|\sum_{k=m+1}^{n}\frac{\sin kx}{k}\right| = \left|\frac{S_n(x)}{n} - \frac{S_m(x)}{m+1} + \sum_{k=m+1}^{n-1} S_k(x) \left(\frac{1}{k} - \frac{1}{k+1} \right)\right| \\ \leqslant \frac{2}{(m+1)|\sin(x/2)|} \\ \leqslant \frac{2\pi}{(m+1)x}\\\leqslant 2\pi$$
since $m = \lfloor 1/x \rfloor$ implies $(m+1)x \geqslant 1$.