Let $0<\alpha$. We wish to examine the continuity of the function $f(x;\alpha)$ as represented by the series
$$f(x;\alpha)=\sum_{n=1}^\infty \frac{\sin(nx)}{n^\alpha}\tag1$$
Dirichlet's test guarantees that for each $\delta>0$, the series in $(1)$ converges uniformly for $x\in [2k\pi+\delta,(2k+1)\pi-\delta]$ for $k\in \mathbb{Z}$. Hence, $f(x;\alpha)$ is continuous on $(2k\pi,(2k+1)\pi)$.
Inasmuch as $f(x;\alpha)$ is odd and $2\pi$-periodic in $x$, it is sufficient, without loss of generality, to test the right-sided continuity of $f(x;\alpha)$ at $x=0$. To that end, we begin with a motivational analysis to provide possible insight.
MOTIVATIONAL ANALYSIS:
The first term in the Euler-Maclaurin Summation Formula for the series in $(1)$ is the integral $I(x;\alpha)$ given by
$$I(x;\alpha)=\int_1^\infty \frac{\sin(xt)}{t^\alpha}\,dt\tag2$$
Enforcing the substitution $t\mapsto t/x$ in $(2)$ reveals for $x>0$
$$x^{\alpha-1}\int_x^\infty \frac{\sin(t)}{t^\alpha}\,dt\tag3$$
We might anticipate, therefore, from $(3)$ that for $x>0$, $f(x;\alpha)$ is $(i)$ continuous at $0$ from the right for $\alpha>1$, $(ii)$ jump discontinuous with jump size $\pi/2$ at $x=0$ from the right for $\alpha=1$, and $(iii)$ unbounded as $x\to 0^+$ for $\alpha <1$.
In the next section, we show that this is indeed the case.
REFINED ANALYSIS
We begin by using summation by parts on the series in $(1)$ to write $f(x;\alpha)$ as
$$\begin{align}
f(x;\alpha)&=\sum_{n=1}^\infty \left(\left(n^{-\alpha}-(n+1)^{-\alpha}\right)\sum_{k=1}^n \sin(kx)\right)\\\\
&=\csc\left(\frac x2\right)\sum_{n=1}^\infty \left(\sin\left(\frac{nx}2\right)\sin\left(\frac{(n+1)x}2\right)\left(n^{-\alpha}-(n+1)^{-\alpha}\right)\right)\tag4
\end{align}$$
Applying the Euler-Maclaurin Summation Formula to the series on the right-hand side of $(4)$, we find
$$\begin{align}
S_N(x;\alpha)&=\sum_{n=1}^N \left(\sin(nx/2)\sin((n+1)x/2)\left(n^{-\alpha}-(n+1)^{-\alpha}\right)\right)\\\\
&=\int_1^N \sin\left(\frac{xt}2\right)\sin\left(\frac{xt}2+\frac x2\right)\left(t^{-\alpha}-(t+1)^{-\alpha}\right)\,dt\\\\
&+\left(1-2^{-\alpha}\right)\sin\left(\frac{x}2\right)\sin\left(x\right)\\\\
&+\int_1^N \frac{d}{dt}\left(\sin\left(\frac{xt}2\right)\sin\left(\frac{xt}2+\frac x2\right)\left(t^{-\alpha}-(t+1)^{-\alpha}\right)\right)\left(t-\lfloor t\rfloor\right)\,dt\tag5
\end{align}$$
Equipped with $(5)$, we are now prepared to analyze the behavior of $f(x;\alpha)$ as $x\to 0^+$. We begin as we did with $(2)$ by enforcing the substitution $t\to t/x$ in the first integral on the right-hand side of $(5)$ and let $N\to \infty$ to find that asymptotically for $x\to 0^+$
$$\begin{align}
I_1(x;\alpha)&=\int_1^\infty \sin\left(\frac{xt}2\right)\sin\left(\frac{xt}2+\frac x2\right)\left(t^{-\alpha}-(t+1)^{-\alpha}\right)\,dt\\\\
&= x^{\alpha-1}\int_x^\infty \sin\left(\frac{t}2\right)\sin\left(\frac{t}2+\frac x2\right)\left(t^{-\alpha}-(t+x)^{-\alpha}\right)\,dt\\\\
&=\alpha x^{\alpha}\int_0^\infty\frac{\sin^2\left(\frac{t}2\right)}{t^{1+\alpha}}\,dt+O(x^{1+\alpha}) \\\\
&=\alpha \left(\frac x2\right)^\alpha \int_0^\infty \frac{\sin^2(t)}{t^{1+\alpha}}\,dt+O(x^{1+\alpha})\\\\
&=\frac12 x^\alpha \int_0^\infty \frac{\sin(t)}{t^\alpha}\,dt+O(x^{1+\alpha})
\end{align}$$
Next, the second term on the right-hand side of $(5)$ is $O(x^2)$.
Finally, it is straightforward to show that the second integral in $(5)$ is $O(x^{1+\alpha})$ as $x\to 0^+$.
CONCLUSION
Putting it all together, we assert that
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{\sin(nx)}{n^\alpha}=x^{\alpha-1}\int_0^\infty \frac{\sin(t)}{t^{\alpha}}\,dt+O(x^\alpha)}\tag6$$
Evidently, $f(x;\alpha)$ is continuous at $0$ when $\alpha>1$, $f(x;\alpha)$ has a discontinuous jump of $\pm \pi/2$ as $x\to 0^\pm$ when $\alpha=1$, and $f(x;\alpha)$ is unbounded as $x\to 0$ when $0<\alpha<1$. This supports the supposition discussed in the Motivational Analysis section.
Best Answer
If: $$ D_N(x) = \sum_{n=1}^{N}\sin(nx) $$ then we have: $$\left| D_N(x) \right|\leq \min\left(N,\frac{1}{|\sin(x/2)|}\right) $$ and the claim follows by partial summation.