TITLE: Uniform convergence of the series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin\frac{1}{nx}$ on $(0,+\infty)$
QUESTION [9 upvotes]: Does the series $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin\frac{1}{nx}$$
converge uniformly on $(0,+\infty)$? First, I guessed that it does, but Abel's test is not applicable here because $\sin\frac{1}{nx}$ is not monotonic, other tools didn't seem to help either. Now I think that there's no uniform convergence, but I can't prove it as well. Any ideas would be appreciated.

REPLY [3 votes]: Here is a proof that there is no uniform convergence.
For any integer $s$ divisible by $2^t$ but not by $2^{t+1}$, let me denote $\tau(s)=s/2^t$. Now consider a sufficiently large integer $m=4u$. I denote $g(m)=\tau(m+1)\cdot\ldots\cdot\tau(2m)$ and $\xi(m)=2g(m)+ \pi m/2$, and also $$\alpha_n(m)=(-1)^{n+1}\sin\frac{\xi(m)}{n}.$$ I claim that $\sum_{n=m+1}^{2m}\alpha_n(m)/n>1/8$, which would imply the lack of uniform convergence.
Note that, for any odd $\mu\in\{m,\ldots,2m\}$, the number $2g(m)/\mu$ is an even integer, so that $\alpha_\mu(m)=\sin(\pi m/(2\mu))\geqslant\sin(\pi/4)>0.7$. Similarly, if $\mu$ divides two but not four, $2g(m)/\mu$ is an odd integer, and again $\alpha_\mu(m)>0.7$. Trivially, $\alpha_\mu(\mu)\geqslant-1$ in other cases.
Thus we have, for any $v\in\{m/4,\ldots,m/2\}$, that $$\sigma_v:=\sum_{n=4v-3}^{4v}\alpha_n(m)/n>\frac{0.7}{4v-3}+\frac{0.7}{4v-2}+\frac{0.7}{4v-1}-\frac{1}{4v}>\frac{2.1}{4v}-\frac{1}{4v}>\frac{1}{4v}.$$ Therefore, $$\sum_{n=m}^{2m}\alpha_n(m)/n=\sum_{v=u+1}^{2u}\sigma_v>\sum_{v=u+1}^{2u}\frac{1}{4v}>\frac{1}{8}.$$