Showing $\sum_{n=0}^{\infty} \frac {\sin(nx)}n $ converges uniformly I have to show that 
$$\sum_{n=0}^\infty \frac {\sin(nx)}n $$ converges uniformly on $[-\pi, -d]\cup[d, \pi]$ but not on $[-\pi, \pi]$
I can't really use Weierstrass's criterion, because the harmonic series diverges.  ANy help?  Thank you.
 A: On the interval $[d, \pi]$ with $d > 0$ we have
$$\left|\sum_{n=1}^m \sin nx\right| = \left|\frac{\sin \left(\frac{mx}{2} \right)\sin \left[\frac{(m+1)x}{2} \right]}{\sin \left(\frac{x}{2}\right)}\right| \leqslant \frac{1}{\sin \left(\frac{d}{2} \right)},$$
and the partial sums of $\sin nx$ are uniformly bounded.
Since $1/n$ converges uniformly to $0$, it follows by the Dirichlet test that the series converges uniformly.  A similar argument can be made for $x \in [-\pi,-d].$
However, the convergence is not uniform on $[-\pi,\pi]$.  This can be proved by showing the partial sums do not satisfy the Cauchy criterion uniformly if $x$ can be chosen arbitrarily close to $0$.  
For any $m \in \mathbb{N},$ let $x_m = \pi/(4m)$.  With $m \leqslant n \leqslant 2m$, we have $\pi/4 \leqslant nx_m \leqslant \pi/2$ and $1/ \sqrt{2} \leqslant \sin n x_m \leqslant 1$. 
Hence,
$$\left|\sum_{n = m}^{2m} \frac{\sin nx_m}{n}\right| \geqslant \frac{1}{\sqrt{2}}\sum_{n=m}^{2m}\frac{1}{n}.$$
Since the harmonic series diverges, the RHS cannot be arbitrarily small, regardless of the choice for $m.$ 
In fact,
$$\lim_{m \to \infty} \sum_{n=m}^{2m}\frac{1}{n} = \log 2$$
