Problem 1.
Show $q$th power of $\sum\frac{\sin n\theta}{n^r}$ (formed by Abel's rule, i.e. $$\nu_n=\sum_{i_1, i_2,\dots,i_q=n} \frac{\sin i_1\theta}{{i_1}^r}\dots\frac{\sin i_q\theta}{{i_q}^r},$$ where $i_j\in\mathbb{Z}_+, r>0, \theta\in\mathbb{R}$) converges when q(1-r)<0.
(When $q=2$, it is similar to Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ converges, and its square (formed by Abel's rule) doesn't.. Show that $\sum\frac{(-1)^{n+1}} {{n}^r} \sum\frac{(-1)^{n+1}} {{n}^s} $ by Abel's rule forms a series that doesn't converge when r+s=1.)
It seems useful to show that $\sum\frac{\sin n\theta}{n^r}$ converges conditionally.
Proof: When $\theta=2k\pi$, the series converges to $0\sum\frac{1}{n^r}=0$.
When $\delta<\theta<2\pi-\delta$, where $\delta$ >0, Write $\theta+\pi$ for $\theta$, then $\sum \sin n\theta=\sum (-)^n\sin n\theta$ and so is bounded i.e. $\exists M$, $\forall p, \sum_{n=1}^p (-)^n\sin n\theta <M$ (Why?). For $\sum\frac{1}{n^r}$ converges to 0 uniformly w.r.t. $\theta$, according to Hardy's test $\sum\frac{\sin n\theta}{n^r}$ converges.
Then how does one prove the convergence of the qth power?
Problem 2.
Show that (1) if $u_n$ is monotonic and $u_n\to0$ as $n\to \infty$, $\delta <\theta<2\pi-\delta$, then $\sum_{n=1}^\infty u_n \cos (n\theta+a)$ is conditionally convergent; (2) if $u_n$ doesn't vanish and is monotonic, $\theta/2\pi$ is rational, then the sum of the series is oscillatory (fluctuating in neighbourhoods of more than one limit point.); when $\theta/2\pi$ is irrational, the sum takes any value between bounds whose difference is $(\lim_{n\to\infty} u_n)\csc\frac{1}{2}\theta$.
(2) could be a consequence of $\sum_{n=1}^\infty \cos (n\theta+a)$ being oscillatory when $\theta/2\pi\in\mathbb{Q}$. The oscillation is suggested in Calculate $\sum_{r=0}^n \cosh(\alpha+2r\beta)$, intuitively, $e^{in\theta}$ is a vector in the real plane , shifting along a chain inscribed in a circle. The chain equals (in the vector addition sense) a smaller vector, which shifts along another chain, which equals a smaller vector. Finally there is a tiny vector shifting along a polygon 'chain', when its 'angle shift' is $2\pi/k$ for some integer k.
If $\theta/2\pi\in\mathbb{R}-\mathbb{Q}$, then similarly the vector vanishes, which however won't orbit a centre along any polygon. Its locus nevertheless becomes smooth and bounded in $\mathbb{C}$, and so $\sum_{n=1}^\infty \cos (n\theta+a)$ varies smoothly and is bounded in $\mathbb{R}$.
Proof:
(1)
$\sum \sin n\theta$ is bounded because
$$\sum_{n=1}^p \sin n\theta=\Im (\sum_{n=1}^p e^{ i n\theta})=\Im (e^{i\theta}\frac{ 1-e^{i p\theta}}{1-e^{i \theta}}),$$ which, for $1-e^{i n\theta}=e^{i0}-e^{i n\theta}=e^{i n\theta/2}(e^{-i n\theta/2}-e^{i n\theta/2})= e^{i n\theta/2}\cdot (-2i)\sin(n\theta/2),$ equals $$\Im (e^{i\theta}\frac{ 1-e^{i p\theta}}{1-e^{i \theta}}) =\Im (e^{i\theta}\frac{e^{i n\theta/2}\cdot (-2i)\sin(n\theta/2)}{e^{i \theta/2}\cdot (-2i)\sin(\theta/2})) =\Im (\frac{e^{i (n+1)\theta/2}\sin(n\theta/2)}{\sin(\theta/2)}) =\frac{\sin[(n+1)\theta/2]\sin(n\theta/2)}{\sin(\theta/2)},$$ bounded by $_-^+$ $\frac{1}{\sin(\theta/2)}.$ (The idea is from an answer.)
$\forall\ p, \sum_{n=1}^p \cos (n\theta+a)$ is bounded by $M$ (independent of p) (consider $\Im(\sum_{n=1}^p e^{i(n\theta+a)})$). Acoording to Hardy's test, $\sum_{n=1}^\infty u_n \cos (n\theta+a)$ converges uniformly.
(2)
When $\theta/2\pi\in \mathbb{Q}$, $\sum_{n=1}^\infty \cos(n\theta+a)$ oscillates among a finite set $H_c$ because:
$\exists$ p, q coprime, such that $\theta=2\pi\frac{p}{q}$ (it suffices to discuss the case when p is odd, i.e. p, 2q coprime), and so
$$\sum_{n=1}^m e^{i (n\theta+a)}=\frac{e^{i [\frac{(m+1)\theta}{2}+a]}\sin(m\theta/2)}{\sin(\theta/2)}
= \frac{\sin(m\frac{2\pi p}{2q})}{\sin(\theta/2)}e^{i (\frac{2\pi (m+1)p}{2q}+a)}
= \frac{\sin(2\pi\frac{mp(\mod 2q)}{2q})}{\sin(\theta/2)}e^{i (2\pi\frac{(m+1)p(\mod 2q)}{2q}+a)},$$
that is, $\sum_{n=1}^\infty e^{i (n\theta+a)}$ oscillates among a finite set $H=\{\frac{\sin(2\pi\frac{1-p}{2q})}{\sin(\theta/2)}e^{i (2\pi\frac{1}{2q}+a)},\frac{\sin(2\pi\frac{2-p}{2q})}{\sin(\theta/2)}e^{i (2\pi\frac{2}{2q}+a)},\dots,\frac{\sin(2\pi\frac{2q-1-p}{2q})}{\sin(\theta/2)}e^{i (2\pi\frac{2q-1}{2q}+a)}\}.$
$\sum_{n=1}^\infty u_n\cos(n\theta+a)$ is oscillatory because:
$\lim_{n\to\infty}u_n=l$ exists, for $u_n$ is decreasing and lower bounded; and $$\sum_{n=1}^\infty u_n\cos(n\theta+a)=\sum_{n=1}^\infty (u_n-l)\cos(n\theta+a)+\sum_{n=1}^\infty l\cos(n\theta+a),$$ where, for $u_n-l\to 0$ steadily, $\sum_{n=1}^\infty (u_n-l)\cos(n\theta+a)$ converges uniformly to $f(\theta),$ and where $\sum_{n=1}^\infty l\cos(n\theta+a)$ oscillates among $lH_c$; and so $\sum_{n=1}^\infty u_n\cos(n\theta+a)$ oscillates among $f(\theta)+lH_c.$
How does one prove the case when $\theta/2\pi\in \mathbb{R}-\mathbb{Q}$? Possibly one needs to calculate the bounds $B_1, B_2$ of limit of the series, and then show that the map $f:m\mapsto \sum_{n=1}^m u_n\cos(n\theta+a), \mathbb{N_+}\to[B_1, B_2]$ is onto.
The question seems to be is $\sum_{n=1}^m (-1)^n \sin n \theta$ bounded for all $m \in \mathbb{N}$?
Note that
$$\sum_{n=1}^m (-1)^n \sin n \theta = \sum_{n=1}^m \cos n\pi \sin n \theta = \sum_{n=1}^m \sin n (\theta+ \pi), $$
and use the well-known result
$$\sum_{n=1}^m \sin nx = \frac{\sin \frac{mx}{2}\sin \frac{(m+1)x}{2}}{\sin \frac{x}{2}}$$
Thus, for all $m \in \mathbb{N}$,
$$\left|\sum_{n=1}^m (-1)^n \sin n \theta \right| = \frac{\left|\sin \frac{m(\theta+\pi)}{2}\right|\, \left|\sin \frac{(m+1)(\theta+\pi)}{2}\right|}{\left|\sin \frac{\theta + \pi}{2}\right|} \leqslant \frac{1}{\left|\sin \frac{\theta + \pi}{2}\right|}$$