Let $\alpha>0$. Show that $\sum_{n=1}^\infty {\sin nx\over n^\alpha}$ converges for all $x\in\Bbb{R}$ and examine continuity of the limit function. First of all I have proved that this series of functions
$$\sum_{n=1}^\infty {\sin nx\over n^\alpha}$$
converges $\forall x\in\Bbb{R}$.
Let $f_n(x)=\sin nx$ and $g_n(x)=\frac{1}{n^\alpha}$.
Let us fix $x\in\Bbb{R}$.
Now, let $a_n=f_n(x)$ and $b_n=g_n(x)$.

*

*Now if $x=2m\pi$ for some $m\in\Bbb{Z}$, $a_n=0$, hence the series $\sum a_nb_n$ converges to $0$.


*Now we assume $x\ne2m\pi$ for any $m\in\Bbb{Z}$.
Then the partial sums of $(a_n)$ to be $A_n=\sum_{k=1}^{n}a_k$. Then
$$ \left| A_n \right| = \left| \sin x + \sin 2x + \cdots + \sin nx \right| = \left|\frac{\sin{\frac{nx}{2}}\sin {\frac{(n+1)x}{2}}}{\sin{\frac{x}{2}}}\right| \le \frac{1}{\left|\sin{\frac{x}{2}}\right|} $$
and $\frac{1}{\left|\sin{\frac{x}{2}}\right|} \in \Bbb{R}$ since $x \ne 2m\pi$.
So, partial sums of $(a_n)$ are bounded. Again $b_n={1\over n^\alpha}$ is monotone decreasing and converges to $0$.
Hence by Dirichlet's test the series $\sum_{n=1}^\infty a_nb_n$ converges.
Thus the series of functions $\sum_{n=1}^{\infty} \frac{\sin nx}{n^\alpha}$ converges for all $x\in\Bbb{R}$.

Let
$$ f(x) = \sum_{n=1}^{\infty} \frac{\sin nx}{n^\alpha} $$
My target is to find where the function $f$ is continuous.
Let us choose a point $a\in\Bbb{R}$ such that $a\ne2m\pi$ for any $m\in\Bbb{Z}$.
Now we can always find a compact interval $I$ containing $a$ such that $\sin \frac{x}{2} \ne 0$ $\forall x\in I$ since we have chosen $a\ne2m\pi$. Since, $I$ is compact interval and $\sin$ function is continuous we have $\min_{x\in I}\lvert \sin \frac{x}{2} \rvert = m > 0 $.
We will show that the series of functions $\sum f_n g_n$ converges uniformly on $I$ by using Dirichlet's test for the series of functions.
Let $(F_n)$ denotes the sequence of partial sums of $(f_n)$.
Then for all $x\in I$, $\left|F_n(x)\right|\le \frac{1}{\lvert\sin\frac{x}{2}\rvert}\le\frac{1}{m}$, hence the sequence of functions $(F_n)$ is uniformly bounded over $I$.
Again, the sequence of function $g_n(x)={1\over n^\alpha}$ is monotone decreasing and converging uniformly to $0$ function.
Thus by Dirichlet's Test for series of functions the series of functions $\sum f_n(x)g_n(x) = \sum_{n=1}^\infty \frac{\sin nx}{n^\alpha}$ converges uniformly over $I$.
Since the functions $f_n g_n$ are continuous on $I$, the limit function $f$ is continuous on $I$.
In particular since $a\in I$, $f$ is continuous at $a$.
So I found $f$ is continuous on the set $\Bbb{R}\setminus \{2m\pi \mid m\in\Bbb{Z}\}$.
But what about the points like $2m\pi$? I observed that $f$ is $0$ at those points.
I am getting no idea how to check continuity at those points.
Can anyone help me in this regard? Thanks for your help in advance.
 A: 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.

