Uniform convergence of a functional series $\sum\limits _{n=1}^{\infty}\frac{x\sin(xn)}{n+x}$ Here's the question: Check series for uniform convergence: $$\sum\limits _{n=1}^{\infty}\frac{x\sin(nx)}{n+x}$$
a) $ x\in E_{1} = (0;\pi) ;$
b) $ x\in E_{2} = (\pi;+\infty) .$
For a) I used the fact that $ \left| \sum\limits_{n=1}^{N}x\sin(nx) \right| \le \left| x\sum\limits_{n=1}^{N}\sin(nx) \right| \le \left| \displaystyle{\frac{x}{\sin(\frac{x}{2})}} \right| \le \pi$ , that means we have bounded partial sums. Sequence $ \{\displaystyle{\frac{1}{n+x}}\} $ is monotone and uniformly goes to 0 for every value of $ x\in E_{1} $ as n goes to $\infty$. Thus, series is umiform convergent on $ E_{1} = (0;\pi) $ . Question is, can I perform the same thing setting partial sums as $ \left| \sum\limits_{n=1}^{N}\sin(nx) \right| $ and sequence as $ \{\displaystyle{\frac{x}{n+x}}\} $ which tells us that it converges to $1$ and finally saying that it converges not uniformly?

For part (b), consider a sequence $x_n = \frac{\pi}{4n} + 2n\pi \in E_2$.  For $n < k \leqslant 2n$ we have
$$\frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}+ 2n^2 \pi \right) = \sin nx_n < \sin k x_n < \sin 2nx_n = \sin \left(\frac{\pi}{2}+ 2n^2 \pi \right) = 1$$
Thus,
$$ \left|\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} \right| =\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} > n \cdot \frac{x_n}{2n + x_n}\cdot \frac{1}{\sqrt{2}} = \frac{2n^2\pi + \frac{\pi}{4}}{\sqrt{2}(2n + 2n\pi + \frac{\pi}{4n})} $$
Since, the RHS tends to $+\infty$ as $n \to \infty$, the series cannot be uniformly convergent on $E_2$, by violation of the uniform Cauchy criterion.
Edit (9/16/2022):
The argument above is flawed because with $x_n = \frac{\pi}{4n} + 2n\pi$, we should have $\frac{\pi}{4} +2n^2\pi < kx_n \leqslant \frac{\pi}{2} + 2n^2\pi + 2n^2\pi$ and we don't have $\sin kx_n > 0$ for all $n< k\leqslant 2n$.
It should be amended as follows.
Consider a sequence $x_n = \frac{\pi}{4n} + 2\pi \in E_2$.  For $n < k \leqslant 2n$ we have $\frac{\pi}{4} < \frac{\pi k}{4n}\leqslant \frac{\pi}{2}$ and
$$\frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}\right) < \sin \frac{\pi k}{4n} =\sin \left(\frac{\pi k}{4n}+ 2k\pi\right) = \sin kx_n  < \sin \left(\frac{\pi}{2} \right) = 1$$
Thus,
$$ \left|\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} \right| =\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} > n \cdot \frac{x_n}{2n + x_n}\cdot \frac{1}{\sqrt{2}} = \frac{2n\pi + \frac{\pi}{4}}{\sqrt{2}(2n + 2\pi + \frac{\pi}{4n})} $$
Since the RHS tends to $\frac{\pi}{\sqrt{2}}\neq 0$ as $n \to \infty$, the series cannot be uniformly convergent ...
