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How can I prove that $$\sum_{n=1}^\infty \frac{\sin(nx)}{\sqrt{n}}$$ converges pointwise on $[-\pi, \pi]$ but not uniformly?

For the pointwise part, I tried to prove it by comparison, using $$\sum_{n=1}^\infty \frac{\sin(nx)}{\sqrt{n}} \leq \sum_{n=1}^\infty \frac{nx}{\sqrt{n}} = \sum_{n=1}^\infty x\sqrt{n},$$ which does not converge.

I also tried $$ \sum_{n=1}^\infty \left| \frac{\sin(nx)}{\sqrt{n}}\right| \leq \sum_{n=1}^\infty \left| \frac{1}{\sqrt{n}}\right|,$$ which does not converge either.

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For the pointwise convergence, use Dirichlet's test.

To see that it's not uniform, note that $\sin(nx) > 2 n x/\pi$ if $0 < n x < \pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum

$$ \eqalign{\sum_{n=1}^N \frac{\sin(n/N)}{ \sqrt{n}} &\ge \frac{2}{\pi N} \sum_{n=1}^N \sqrt{n} \cr&\ge \frac{2}{\pi N} \int_0^N \sqrt{t}\; dt = \frac{4\sqrt{N}}{3\pi}}$$

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Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $\lim_{x\to 0^+}\sum_{n\geq 1}\frac{\sin(n x)}{\sqrt{n}}$ through a convolution with an approximate identity:

$$\begin{eqnarray*}\lim_{x\to 0^+}\sum_{n\geq 1}\frac{\sin(n x)}{\sqrt{n}}&=&\lim_{m\to +\infty}\sum_{n\geq 1}\frac{1}{\sqrt{n}}\int_{0}^{+\infty}m\sin(nx) e^{-mx}\,dx\\&=&\lim_{m\to +\infty}\sum_{n\geq 1}\frac{1}{\sqrt{n}}\cdot\frac{mn}{m^2+n^2}.\end{eqnarray*}$$ The RHS is $+\infty$ since by Riemann sums $$ \lim_{m\to +\infty}\sum_{n\geq 1}\frac{\sqrt{mn}}{m^2+n^2}=\int_{0}^{+\infty}\frac{\sqrt{x}}{1+x^2}\,dx = \frac{\pi}{\sqrt{2}}.$$ Much simpler, once $f(x)$ is defined as $\sum_{n\geq 1}\frac{\sin(nx)}{\sqrt{n}}$ we have $$ \int_{-\pi}^{\pi}f(x)^2\,dx = \pi\sum_{n\geq 1}\frac{1}{n} = +\infty $$ by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-\pi,\pi]$.

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