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.
 A: 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
$$\sum_{n=m}^{3m} \frac{1}{n}\sin \frac{x_m}{n} \geqslant \frac{1}{2} \sum_{n=m}^{3m}\frac{1}{n} > \frac{1}{2} \log 3.$$
A: Let $\:t_n=1/n,\:\:u_n=\sin\left(\large\frac{x}{n}\normalsize\right)\:$ and $\:S_n(x)=\sum_{j=1}^nt_ju_j,\quad \underbrace{(j,n)\in\mathbb N^2,\:\:j\le n.}_{\text{Property}\{\star\}} $
Therefore, $$S_n(x)=\sum_{j=1}^nt_j(U_j-U_{j-1}),\quad\:U_j=\sum_{k=1}^ju_k\:\{\star\}.\\\implies S_n(x)=t_nU_n+\sum_{j=1}^{n-1}t_{j}U_j-\sum_{j=1}^{n-1}t_{j+1}U_{j-1}=t_nU_n+\sum_{j=1}^{n-1}(t_j-t_{j+1})U_j.$$
Now, $$U_j=\sum_{k=1}^n\sin(x/k)=\mathcal{Im}\left(\sum_{k=1}^ne^{ix/k}\right)=\mathcal{Im}\left(\frac{e^{ix}-e^{ix/(n+1)}}{1-e^{ix}}\right)=\mathcal{Im}\left(\frac{e^{ix}-e^{ix/(n+1)}}{-2i\sin\left(\frac{x}{2}\right)e^{ix/2}}\right).$$
$\implies\forall x\in\{0,2\pi\mathbb Z\},\:\:S_n(x)\:$ becomes unbounded as $\:n\to\infty$.
A: 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 $\:\mathbb R_+.$
A: For $x=n\pi /2$,the series reduces to harmonic series that does not even converge.So the series cannot converge uniformly on $\mathbb R$.
