$\sum_{n=1}^{\infty} \frac {\sin(nx)}{\sqrt{n}}$ uniform convergence in $[0,2\pi]$ Let $\sum_{n=1}^{\infty} \frac {\sin(nx)}{\sqrt {n}}$ be a series of functions.
How can I show that it is not uniformly convergent in $[0,2\pi]$?
I thought about the series $\sum_{n=1}^{\infty} \frac {\sin(n\pi/2)}{\sqrt {n}}$, which is in range but non convergence. Is that enough to disprove it?
Is there a more formal way?

Notice that
$$\sum_{n\geq 1}\frac{\sin(n\pi/2)}{\sqrt{n}}$$
is convergent by Dirichlet's test. However, uniform convergence would imply convergence in $L^2$, and the $L^2$ norm of 
$$\sum_{n=1}^{N}\frac{\sin(nx)}{\sqrt{n}}$$
grows unbounded. As an alternative, you may prove that
$$ f(x)=\sum_{n\geq 1}\frac{\sin(nx)}{\sqrt{n}} $$
is neither bounded or continuous on $(0,2\pi)$. However, this is way more difficult than proving that $f$ is not square-integrable.
