Does $\sum_{n\ge1}\frac1n\sin\left(\frac xn\right)$ converges uniformly? On each bounded interval $[a,b]$ : 
$\left|\frac1n \sin\left(\frac xn\right)\right|\le \frac{\max\{|a|,|b|\}}{n^2}$, 
the series $\sum_{n\ge 1}\frac {\max\{|a|,|b|\}}{n^2}$ converges,

therefore
$\sum_{n\ge1}\frac1n \sin\left(\frac xn\right)$ converges uniformly by the Weierstrass M-Test.

May I conclude from this statement that this series converges uniformly on $\mathbb{R}$?

Thanks.
The series does not converge uniformly on $\mathbb{R}$.

Consider $x_m = m\cdot \frac{\pi}{2}$ for $m \in \mathbb{Z}^+$. Then

$$\sin \frac{x_m}{n} \geqslant \frac{1}{2}$$

for $m \leqslant n \leqslant 3m$, and so

$$\sum_{n=m}^{3m} \frac{1}{n}\sin \frac{x_m}{n} \geqslant \frac{1}{2} \sum_{n=m}^{3m}\frac{1}{n} &gt; \frac{1}{2} \log 3.$$
Choose a $\:x\:$ that will create a contradiction such as $\:x=n.$

$$\implies \sum_{n\in\mathbb N}\frac{1}{n}\sin(1)\overset{\text{Harmonic}}{=}\infty$$
So that it does not converge uniformly on any bounded set since it doesn not converge on $\:\mathbb R_+.$