Find an example of a sequence not in $l^1$ satisfying certain boundedness conditions. This question is about getting a concrete example for this question on bounded holomorphic  functions posed by @user122916 (something that he really expected as explained in the comments).
Give an example of a sequence of complex numbers $(a_n)_{n\ge 0}$ so that 
\begin{eqnarray}
|\sum_{n\ge 0} {a_n z^n} | &\le &1  \text{ for all }z \in \mathbb{C}, |z| < 1 \\
\sum_{n\ge 0} |a_n| &=& \infty
\end{eqnarray}
Such sequences exist because there exist bounded holomorphic functions on the unit disk that do not have a continuous extension to the unit circle ( one finds a bounded Blaschke product with zero set that contains the unit circle in its closure). However, a concrete example escapes me. Note that all this is part of the theory of $H^{\infty}$ space, so the specialists might have one at hand.

An example is $f(z) = \exp {(-\frac{1+z}{1-z})}.$ As $-\frac{1+z}{1-z}$ is a conformal map of the open unit disc $\mathbb {D}$ onto the left half plane, $f$ is bounded and holomorphic in $\mathbb {D}.$ We have $f$ continuous on $\overline {\mathbb {D}} \setminus \{1\}.$ Check that on $\partial \mathbb {D}\setminus \{1\},$ we have $|f| = 1.$ However $f(r) \to 0$ (to say the least) as $r\to 1^-.$ It follows that $f$ does not have a continuous extension to $\overline {\mathbb {D}}.$ Hence for this $f, \sum_{n=1}^{\infty}|a_n| = \infty.$
