Does this series converge uniformly $$f(x) =\sum_{n=1}^\infty \frac{\sin (\frac x n ) }{n}$$ on $\Bbb R$. If $n$ is fairly large enough, we get $\sin (\frac x n ) \le \frac x n$ but still it depends on value of $x$ so choice of $n$ may depend on $x$.
Added: Looks like there is workaround from $f'(x)$ which can be shown differentiable which implies $f(x)$ is continuous as well as the series is uniformly convergent.
The convergence is not uniform for $x \in \mathbb{R}$. The uniform convergence of the series of derivatives only implies uniform convergence of the original series on finite intervals.
For any $m \in \mathbb{N},$ take $x_m = (m\pi)/2$.
With $m < n \leqslant 2m$, we have $1/(2m) \leqslant 1/n < 1/m$ and $\pi/4 \leqslant x_m/n \leqslant \pi/2$. This implies that $1/ \sqrt{2} \leqslant \sin (x_m/n)\leqslant 1$.
Hence,
$$\left|\sum_{n = m+1}^{2m} \frac{\sin \left(\frac{x_m}{n}\right)}{n}\right| \geqslant \frac{m}{2m\sqrt{2}}= \frac{1}{2\sqrt{2}}$$
The RHS cannot be arbitrarily small, regardless of the choice for $m$ -- violating the Cauchy criterion for uniform convergence.