Uniform convergence of $\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}\sin(nx+\frac{1}{nx}), \alpha > 0$ in $[\delta,2\pi-\delta]$ The solution says using Dirichlet's test on $\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}\sin(nx+\frac{1}{nx})$.
$x\in[\delta,2\pi-\delta],\delta$ is a sufficiently small positive real number.
The question is that how to prove $\sum_{n=1}^{m}\sin(nx+\frac{1}{nx})$ is bounded?
 A: It has been 3 months since I posted this question.
The main idea is from: https://math.stackexchange.com/q/568079
We have
$$
\left|\frac{1}{n^\alpha}\sin(nx+\frac{1}{nx})-\frac{1}{n^\alpha}\sin(nx)\right|
<\frac{1}{n^\alpha}\left|\sin(nx+\frac{1}{nx})-\sin(nx)\right|<
\frac{1}{n^\alpha}\frac{1}{nx}
$$
for all $x\in[\delta,2\pi-\delta],\ n\in\mathbb{N}$.
Since $\sum \frac{1}{n^\alpha}\frac{1}{nx}$ converges ($1+\alpha >1$) uniformly on $x\in[\delta,2\pi-\delta]$,
Using Cauchy sequence we get: for all $\epsilon > 0$
$$
\sup_{x\in[\delta,2\pi-\delta]}\left|\sum_n^m \frac{1}{n^\alpha}\frac{1}{nx}\right| = \sup_{x\in[\delta,2\pi-\delta]}\sum_n^m \frac{1}{n^\alpha}\frac{1}{nx} < \epsilon \tag{1}
$$ for sufficiently large $m,n$.
Using Dirichlet's test, $\sum\frac{1}{n^\alpha}\sin(nx)$ converges uniformly so:
$$
\sup_{x\in[\delta,2\pi-\delta]}\left|\sum_n^m \frac{1}{n^\alpha}\sin(nx)\right| < \epsilon \tag{2}
$$
use (1) and (2) we get:
$$
\sum_n^m \frac{1}{n^\alpha}\sin(nx+\frac{1}{nx})>\sum_n^m \left(\frac{1}{n^\alpha}\sin(nx)-\frac{1}{n^\alpha}\frac{1}{nx}\right) > 
\sum_n^m \frac{1}{n^\alpha}\sin(nx)-\sum_n^m \frac{1}{n^\alpha}\frac{1}{nx} > -2\epsilon
$$
and
$$
\sum_n^m \frac{1}{n^\alpha}\sin(nx+\frac{1}{nx})<\sum_n^m \left(\frac{1}{n^\alpha}\sin(nx)+\frac{1}{n^\alpha}\frac{1}{nx}\right) < 
\sum_n^m \frac{1}{n^\alpha}\sin(nx)+\sum_n^m \frac{1}{n^\alpha}\frac{1}{nx} < 2\epsilon
$$
for all $x\in[\delta,2\pi-\delta]$, so $\sum \frac{1}{n^\alpha}\sin(nx+\frac{1}{nx})$ converges uniformly.
