# converge pointwise but not uniformly

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.

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}}

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 $$[-\pi,\pi]$$.