Functions in $\mathcal{D}$ are finite test functions in $C^\infty(\mathbb{R})$
$\mathcal{D}'$ are distributions (genralized functions)
Do I have to check that $\forall \phi \in \mathcal{D}$: $$\lim_{\epsilon \to 0} \int_\epsilon^a\sum\limits_{n=1}^\infty \frac{e^{inx}}{n}\phi(x)\mathrm{d}x < \infty,$$ or what?
The Fourier series for $f(x)=-\ln(1-e^{ix})$ converges uniformly to $f$ for $x \in [\delta,2\pi-\delta]$ for any fixed $\delta > 0$. For $n \ne 0$, the Fourier coefficient $c_n$ satisfies \begin{align} 2\pi c_n & =\int_{0}^{2\pi}e^{-inx}f(x)dx \\ & =\frac{e^{-inx}-1}{-in}f(x)|_{x=0}^{2\pi}-\int_{0}^{2\pi}\frac{e^{-inx}-1}{-in}f'(x)dx \\ & = -\int_{0}^{2\pi}\frac{e^{-inx}-1}{in}\frac{ie^{ix}}{1-e^{ix}}dx. \end{align} For $n > 0$, $e^{inx}-1=(e^{ix}-1)(1+e^{ix}+e^{2ix}+\cdots+e^{i(n-1)x})$. So you can directly determine all of the Fourier coefficients, and you find that--as expected--the Fourier series for $f$ is $\sum_{n=1}^{\infty}\frac{e^{inx}}{n}$. By Abel's theorem for power series, $$ \lim_{r\uparrow 1}\sum_{n=0}^{\infty}\frac{r^{n}e^{inx}}{n}=f(x), $$ and the convergence is uniform in $r$ for $x\in [\delta,2\pi-\delta]$ if $\delta > 0$. That also gives you some uniform bounds, knowing the power series $F(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n}$.