Prove that $\sum_{n=1}^\infty \frac{e^{inx}}{n}$ is uniformly convergente by variable change

Question

Prove that $$\sum_{n=1}^\infty \frac{e^{inx}}{n}$$ is uniformly convergente in $[\delta,2\pi-\delta]$ with $\delta \in(0,\pi)$

It could be proved by the  Dirichelt's criterion; but if we want to solve it by variable change, I get a problem.

If we denote $\omega = e^{ix}$, then $$ \sum_{n=1}^\infty \frac{e^{inx}}{n}=\sum_{n=1}^\infty \frac{\omega^n}{n}$$ that is convergente in $\bar D(0,\delta)$ with $\delta<1$. How I can see the radio is of convergence for $x$?

Answer 1

It may be enlightening to switch to a complex variable. Thus let $z\in \mathbb{C}$ with $Imz>0$. Then \begin{equation*} I(z)=\sum_{n=1}^{\infty }\frac{1}{n}e^{inz} \end{equation*} converges and \begin{equation*} \partial _{z}I(z)=\sum_{n=1}^{\infty }(e^{iz})^{n}=\frac{e^{iz}}{1-e^{iz}} \end{equation*} We see that this expression is analytic except in the poles, defined by \begin{equation*} e^{iz}=e^{i(z+k2\pi )}=1\Rightarrow z_{k}=k2\pi ,\;k=0,\pm 1,\pm 2,\cdots \end{equation*} Formally \begin{equation*} I(z)=\ln (1-e^{iz}) \end{equation*} but we have to be careful with the logarithm. Outside the poles we can let $ Imz$ tend to $0$, in particular for $x=Rez\in (0,2\pi )$

Answer 2

The series converges uniformly for $|\omega|\le\delta<1$. You can't recuperate an interval of convergence for $x$ since $|e^{ix}|=1$.