Let $a_n$, a sequence monotonically decreasing to $0$. Consider
$$\sum_{n=1}^\infty a_n\sin(nx)$$
Is the series converges uniformly on $[\varepsilon, 2\pi-\varepsilon]$? ($\varepsilon > 0$)
Basically we could use Dirichlet's test. We want to show that $\sum_{n=1}^\infty \sin (nx)$ is bounded. Indeed:
$$\sum_{n=1}^\infty \sin (nx) = \frac{i}{2}\left( \sum_{n=1}^\infty (e^{ix})^n + \sum_{n=1}^\infty (e^{-ix})^n\right) \le \frac{i}{2} \left( \frac{1}{1-e^{ix}} +\frac{1}{1-e^{-ix}} \right) \le \frac{1}{1-e^{i(2\pi-\varepsilon)}} <\infty$$
BUT, clearly, $$g(\frac{\pi}{2}) = \sum_{n=1}^\infty \sin \frac{n\pi}{2} = \infty$$
Where is the mistake?
You last "clearly" is problematic. Because $g(\pi/2)$ is bounded between 0 to 1.