$S(x)=\sum_{n=1}^{\infty}a_n \sin(nx) $, $a_n$ is monotonic decreasing $a_n\to 0$: Show uniformly converges within $[\epsilon, 2\pi - \epsilon]$ $S(x)=\sum_{n=1}^{\infty}a_n \sin(nx) $, $a_n$ is monotonic decreasing $a_n\to 0$, when ${n \to \infty}$. I need to prove that for every $\epsilon >0$, the series is uniformly converges within $[\epsilon, 2\pi - \epsilon]$. Can I use Dirichlet and say that $\sum_{0}^{M} \sin (nx)< M$ for every x in the interval and since $a_n$ is uniformly converges to 0 ( uniformity since it does not depend on $x$), so the series is uniform convergent in this range?
In addition I need to prove that if $\sum_{n=1}^{\infty} a_n^2 = \infty$  so the series is not uniform convergent in $[0, 2 \pi]$, Since I know that from $n_0$ and on $a_n^2< a_n$ I used an inequality, again I'm not sure of that.
In the other hand, maybe I need to use Fourier series somehow. 
Thanks for the help!

Indeed, the fact that the sequence $\{\sum_{n=1}^N\sin(nx)\}$ is bounded will help us. More precisely, fix $\varepsilon>0$. 


*

*Show that we can find a constant $C=C(\varepsilon)$ such that for all $x\in [\varepsilon,2\pi-\varepsilon]$ we have $\left|\sum_{n=1}^N\sin(nx)\right|\leq C$ (to do that compute the sum). 

*Now we use Abel's tranform. We denote $s_N(x):=\sum_{n=0}^N\sin(nx)$ and $S_N(x):=\sum_{n=1}^Na_n\sin(nx)$. Then show that 
$$S_N(x)=\sum_{n=1}^Na_n(s_n(x)-s_{n-1}(x))=a_Ns_N(x)-a_1s_0(x)+\sum_{n=1}^{N-1}(a_n-a_{n-1})s_n(x).$$

*Prove that the first two terms gives a uniformly convergent sequence (to $0$) whereas the third is a normally convergent sequence on $[\varepsilon,2\pi-\varepsilon]$, hence an uniformly convergent sequence on this interval. 


For the second question, I will give you some steps:


*

*Compute for fixed integers $m$ and $n$ $\int_0^{2\pi}S_n(x)S_m(x)dx$. 

*If we assume that $\{S_N\}$ is uniformly convergent on $[0,2\pi]$, then 
$$\lim_{n,m\to \infty}\int_0^{2\pi}S_n(x)S_m(x)dx=\int_0^{2\pi}S(x)^2dx.$$

*Conclude. 

