# Uniform convergence of $\sum_{k=2}^\infty \frac{\sin(kx)}{k \ln(k)}$?

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.

• Dirichlet test. The other series is also tested by this. – xbh Oct 26 '18 at 23:47

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 $$(m+1)x \geqslant 1$$.

• Thank you. I didn't expect uniform convergence every where! I thought it would be like the other example. Could you explain further about bounding the other sum. – WoodWorker Oct 27 '18 at 0:09
• @WoodWorker: OK -- I'll add some more detail. – RRL Oct 27 '18 at 0:18