Uniform convergence of the series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin\frac{1}{nx}$ on $(0,+\infty)$ 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.
 A: 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, $$\sum_{n=m}^{2m}\alpha_n(m)/n=\sum_{v=u+1}^{2u}\sigma_v>\sum_{v=u+1}^{2u}\frac{1}{4v}>\frac{1}{8}.$$
A: this is an alternate series, and for a fixed x, we can find an No such as $\frac{1}{n*x}$ is always smaller than pi/2 (I want the monotony of sin), if n>=No.
So if we write the general term $(-1)^\left(n+1\right)*Vn(x)$, we have:
(Vn(x)) decreases, Vn(x)-->0, when n--> ∞, so the theorem about the rest of alternate series can be used here:
The rest verify: |Rn(x)| ~ |Vn(x)|, with Rn(x) = $\sum_{k=n+1}^\infty Vk(x)$
So |Rn(x)| ~ $\frac{1}{x*n^2}$ , with the inequality : |Rn(x)| =< $|V_n(x)|$ =< $\frac{1}{x*n^2}$
So you have the uniform convergence on every interval of the type [A;+∞[, but not on [0;+∞[
