How to prove that $\sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ is discontinuous at $0$? 
How to prove that $\sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ is discontinuous at $0$?

As we know, $\sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ converges uniformly on $[\delta,\pi-\delta]$ for $\delta>0$. and not uniformly on $(0,\pi)$.
But this could not be used to show discontinuity of $\sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}=S(x)$ at $0$. Any ideas?
 A: To establish the discontinuity of $S(x)$ at $x = 0$, it suffices to show that $S(\pi/N) \to \infty$ as $N \to \infty$. Indeed, by grouping the terms into $N$ consecutive terms, the defining sum for $S(\pi/N)$ is recast as
\begin{align*}
S(\pi/N)
&= \sum_{r=1}^{N} \sin\left(\frac{r\pi}{N}\right) \sum_{q=0}^{\infty} \frac{(-1)^q}{\sqrt{qN + r}}.
\end{align*}
Then by the property of the alternating series, we have
$$ \frac{1}{\sqrt{r}} - \frac{1}{\sqrt{N + r}} \leq \sum_{q=0}^{\infty} \frac{(-1)^q}{\sqrt{qN + r}} \leq \frac{1}{\sqrt{r}}. $$
Together with the fact that $\sin(r\pi/N) \geq 0$ for all $ r = 1, 2, \ldots, N$, we obtain the bound
\begin{align*}
S(\pi/N)
&\geq \sum_{r=1}^{N} \sin\left(\frac{r\pi}{N}\right) \left( \frac{1}{\sqrt{r}} - \frac{1}{\sqrt{N + r}} \right) \\
&= \sqrt{N} \cdot\sum_{r=1}^{N} \sin\left(\frac{r\pi}{N}\right) \left( \frac{1}{\sqrt{r/N}} - \frac{1}{\sqrt{1 + r/N}} \right) \frac{1}{N}.
\end{align*}
Now the sum in the last line, excluding the prefactor $\sqrt{N}$, is the Riemann sum for the integral
$$ C := \int_{0}^{1} \sin(\pi x) \left( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{1+x}} \right) \, \mathrm{d}x > 0, $$
and so, it follows that
$$ S(\pi/N) \geq (C + o(1)) \sqrt{N}. $$
This shows that $S(\pi/N) \to \infty$ as $N \to \infty$, hence $S(x)$ is discontinuous at $x = 0$.
