$\sum_{n=1}^{\infty} \frac {\sin(nx)}{\sqrt{n}}$ 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?
 A: 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.
A: $$\:\sin (nx)>\frac{\:1}{\sqrt2}\:,\:\:\:\forall x\in\left]\frac{2\pi}{n}(j+\frac{1}{4}),\frac{2\pi}{n}(j+\frac{3}{4})\right[=I_{j,n}\:\:\:\:(j,n)\in\mathbb N^2.$$
$$\implies \sum_{n\in\mathbf N}{\sin(nx)\over  n^{\xi}}>\sum_{n\in\mathbf N}{1\over \sqrt {2}n^{\xi}}=\infty\quad \forall\xi\in\:]0,1],\:\forall x\in I_{j,n}.$$
We can see $\: I_{j,n}\:$ as $\:x\:$ satisfying $\:\:{\large{\pi\over4n}}<x-{\large{2\pi j\over n}}<{\large{3\pi\over4n}}.$
Else, we can use the Abel transform: $$\mathcal H_n=\sum_{k=0}^na_kb_k,\:\:B_n=\sum_{k=0}^nb_k,\quad \forall k\le n\:\:a_k\in\mathbb R,\:b_k\in\mathbb C.\\\implies\mathcal H_n=a_nB_n-\sum_{k=0}^{n-1}(a_{k+1}-a_{k})B_k $$
In our case, $\:a_k=k^{-\xi}\implies\forall\xi\in\:]0,1]\quad \sum (a_{k+1}-a_k)\:$ is telescoping.
Also, $\:B_k=\sum_{j=0}^k\sin(jx)=\mathcal Im\left(\sum_{j=0}^ke^{ji\phi} \right)\quad j\le k,\:i\in \mathbb C,\:\phi\in \mathbb R$
Now,$$B_k=\mathcal Im \left(\sum_{j=0}^ke^{ji\phi} \right)=\mathcal Im\left({1-e^{i\phi(n+1)}\over 1-e^{i\phi}}\right)=\mathcal Im\left({e^{ni\phi}\over e^{i\phi/2}}{e^{-ni\phi}-e^{i\phi}\over e^{-i\phi/2}-e^{i\phi/2}}\right)\\=\mathcal Im\left({e^{i\phi(n+1)}-1}\over 2i\sin(\phi/2)e^{i\phi/2} \right).$$
This is why whenever $\:\phi\in\{0,2\pi\mathbb Z\},\:\mathcal H_n\:$ becomes unbounded.
