Let $\phi$ be defined as such:
\begin{equation} \phi(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2} \end{equation}
By Hadamard's theorem, the radius of convergence of $\phi$ equals $1$. In particular for $z \in \partial D(0, 1)$, the series remains absolutely convergent (it converges to $\frac{\pi^2}{6}$) therefore $\phi$ is absolutely convergent over $\overline{D(0, 1)}$.
I now want to prove that for all $\varepsilon > 0$, one can find a $n_\varepsilon >0 $ such that, for all $ m \geq n_\varepsilon$,
\begin{equation} \Big|\sum_{n \geq m}^{\infty} \frac{z^n}{n^2} \Big| < \varepsilon, \quad z \in D(0, 1). \end{equation}
I'm only looking for tips on how to get started. Thank you.