Does $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$ in $[0,2\pi]$

This question arises because of a problem I was doing (Bartle 3rd edition, section 9.4 problem 3). It was like this.
Given $a_n$ a decreasing sequence of positive numbers and suppose that $$\sum_{n=0}^{\infty}{a_n \sin{(nx)}}$$ Converge uniformly (It doesn't specify the domain, so I guess is for every x). Prove that $n a_n \to 0$.
Clearly $\frac{1}{n}$ fits the description of $a_n$, and $n \frac{1}{n} \to 1 \neq 0$, so this would prove that there is a mistake in the problem if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$.
So my question is if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for every $x$.

(I know that the series converge uniformly for every x in $[\delta, 2\pi - \delta]$, for $0 < \delta <2\pi$ by using the Dirichlet criterion.)

Solution 1:

The sum of the series is non-continuous (you can view this as the Fourier series for a saw-tooth function; or just check the behavior around x=0), so the convergence cannot be uniform. Each summand is obviously a continuous function, and a uniformly convergent series of continuous functions is continuous.

Solution 2:

Hint:

Use Cauchy Criterion to prove that your infinite series isn't uniformly converges for all $x\in[0,2\pi]$.