Show that $\sum_{n=1}^\infty \frac{\sin (nx)}{n}$ converges uniformly on $[\pi /2 , 3\pi /2]$ I must show that $\sum_{n=1}^\infty \frac{\sin (nx)}{n}$ converges uniformly on $[\pi /2 , 3\pi /2]$ - I know that it holds on $[\pi /2 , \pi ]$, which should also be sufficient as a proof, but I don't know how to prove it.
 A: We want to be able to use Dirichlet uniform convergence test
So we need that
i)
1/n is monotonous decreasing function
ii)
1/n uniformly converges to 0 (which it does since it doesn't depend on variable x)
iii)  
$$
(\exists M>0)(\forall n\in \mathbb{N})(\forall x\in [\pi/2,3\pi/2])
$$
$$
\left | \sum_{m=1}^{n} \sin(mx)  \right | \le M
$$
We can prove that by 
$$
\left | \frac{1}{\sin{\frac{x}{2}}}\sum_{m=1}^{n} \sin{mx}\sin{\frac{x}{2}}  \right | = \left | \frac{1}{2\sin{\frac{x}{2}}}\sum_{m=1}^{n} (\cos{\frac{(2m-1)x}{2}}-\cos{\frac{(2m+1)x}{2}})  \right | =
$$
$$
\left | \frac{\cos{\frac{x}{2}}-\cos{\frac{(2n+1)x}{2}}}{2\sin{\frac{x}{2}}} \right | \le \left | \frac{1}{\sin{\frac{x}{2}}} \right |
$$
So when $x\in [\pi/2,3\pi/2])$ we know $\frac{x}{2} \in [\pi/4,3\pi/4]$
$$
\max_{x \in [\pi/4,3\pi/4]} \left | \frac{1}{\sin{\frac{x}{2}}} \right | = \frac{2}{\sqrt{2}}
$$
Hence using Dirichlet uniform convergence test we conclude that beginning function series uniformly converges on $[\pi/2,3\pi/2]$ 
