Determine whether the series converges uniformly How to determine whether the series $\sum\limits_{n=1}^\infty\frac{\sqrt x\sin nx}{n}$ is uniformly convergent on $(0,\pi)$ or not? Using the Cauchy criterion for uniform convergence, I can show that series $\sum\limits_{n=1}^\infty \frac{\sin nx}{n}$ does not converge uniformly on $(0,\pi)$, hence the sequence $S_n=\sum\limits_{k=1}^n \sin kx$ is not uniformly bounded on $(0,\pi)$. But I fail to apply Cauchy criterion to the series in question. To me it seems that it is not uniformly convergent, because I could only bound the sequence of partial sums $S_n=\sum\limits_{k=1}^n \sqrt{x}\sin kx$ above with the function $\frac{\sqrt{x}}{sin(\frac x 2)}$ which tends to infinity as $x$ goes to zero, but still I don't see how to approach this problem.
 A: $\displaystyle\sum_{n=1}^\infty\frac{\sqrt x\sin nx}{n}$ converges pointwise to $\displaystyle\frac{\sqrt x(\pi-x)}{2}$ on $[0,\pi]$. Moreover $\displaystyle\sum_{n=1}^\infty\frac{\sin nx}{n}$ is bounded. Given $\epsilon>0$ choose $\delta>0$ such that
$$
\Bigl|\sum_{n=1}^\infty\frac{\sqrt x\sin nx}{n}\Bigr|\le\epsilon,\quad\frac{\sqrt x(\pi-x)}{2}\le\epsilon,\quad 0\le x\le\delta.
$$
Since $\displaystyle\sum_{n=1}^\infty\frac{\sin nx}{n}$ converges uniformly to $\displaystyle\frac{\pi-x}{2}$ on $[\delta,\pi]$, we can choose $n_0$ such that if $n\ge n_0$ then
$$
\Bigl|\sum_{n=1}^\infty\frac{\sqrt x\sin nx}{n}-\frac{\sqrt x(\pi-x)}{2}\Bigr|\le\epsilon,\quad\delta\le x\le\pi.
$$
Putting it all together we see that the convergence is uniform on $[0,\pi]$.
Proof of the uniform boundedness of the partial sums.
Using Abel's summation formula show that
$$
\Bigl|\sum_{k=j}^n\frac{\sin(k\,x)}{k}\Bigr|\le\frac{2}{j\sin(x/2)}.
$$
Now
$$\begin{align}
\Bigl|\sum_{k=1}^\infty\frac{\sin(k\,x)}{k}\Bigr|&\le x\Bigl|\sum_{1\le k\le 1/x}\frac{\sin(k\,x)}{k\,x}\Bigr|+\Bigl|\sum_{1/x<k\le n}\frac{\sin(k\,x)}{k}\Bigr|\\
&\le1+\frac{2\,x}{\sin(x/2)}\\
&\le 1+2\,\pi.
\end{align}$$
A: Here is my unfinished attempt to write down the solution suggested by Julián. Hopefully someone can check if my argument is complete and correct it.
1) Since $\lim\limits_{x\rightarrow 0+} \frac{\sqrt{x}(\pi-x)}{2}=0$, we have that $\forall \epsilon>0 \space \exists \delta_{\epsilon}>0 : \forall x\in (0,\delta_{\epsilon}) \implies \left|\frac{\sqrt{x}(\pi-x)}{2}\right|<\epsilon$.
2) Since $S_n=\sum\limits_{k=1}^n\frac{\sqrt{x}\sin kx}{k}$ converges pointwise to $\frac{\sqrt{x}(\pi-x)}{2}$ on $(0,\delta_{\epsilon})\subset(0,\pi)$, we have that $\forall \epsilon>0 \space \forall x\in(0,\delta_{\epsilon}) \space \exists N_{x,\epsilon}: \forall n>N_{x,\epsilon} \implies \left|\sum\limits_{k=1}^n\frac{\sqrt{x}\sin kx}{k}-\frac{\sqrt{x}(\pi-x)}{2}\right|<\epsilon \implies$ $\epsilon>\left|\sum\limits_{k=1}^n\frac{\sqrt{x}\sin kx}{k}\right|-\left|\frac{\sqrt{x}(\pi-x)}{2}\right| \implies \left|\sum\limits_{k=1}^n\frac{\sqrt{x}\sin kx}{k}\right|<\left|\frac{\sqrt{x}(\pi-x)}{2}\right|+\epsilon<\epsilon+\epsilon=2\epsilon.$
Now suppose we are given $\epsilon>0$, then from point 1) above we have that $\exists \delta_{\epsilon}:$ $\sup\limits_{x \in (0,\delta_{\epsilon})}\left|\sum\limits_{k=1}^n\frac{\sqrt{x}\sin kx}{k}-\frac{\sqrt{x}(\pi-x)}{2}\right|\leqslant \sup\limits_{x \in (0,\delta_{\epsilon})}\left(\left|\sum\limits_{k=1}^n\frac{\sqrt{x}\sin kx}{k}\right|+\left|\frac{\sqrt{x}(\pi-x)}{2}\right|\right)$ $\leqslant$ $\sup\limits_{x \in (0,\delta_{\epsilon})}\left|\sum\limits_{k=1}^n\frac{\sqrt{x}\sin kx}{k}\right|+\sup\limits_{x \in (0,\delta_{\epsilon})}\left|\frac{\sqrt{x}(\pi-x)}{2}\right|\leqslant$
$b(n)+\epsilon$,
where $b(n)=\sup\limits_{x \in (0,\delta_{\epsilon})}\sum\limits_{k=1}^n\frac{\sqrt{x}\sin kx}{k}$ exists since both $\sqrt{x}$ and $\sum\limits_{k=1}^n\frac{\sin kx}{k}$ are bounded on $(0,\delta_{\epsilon})$.
Here I have a problem showing that $\lim\limits_{n\to\infty}b(n)=0$, because for each $n$ supremum can be attained at a different point $x_{0}(n)\in(0,\delta_{\epsilon})$. And pointwise convergence gives us different $N_{x,\epsilon}$ for each $x$.  
