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.

Consider a Blaschke product $$ f(z) = \prod_{n=1}^\infty \dfrac{z - r_n}{1 - r_n z}$$
where $0 \le r_n < 1$ and $r_n \to 1$ as $n \to \infty$;  this converges in the open unit disk if $\sum_n (1 - r_n) < \infty$.  If its Maclaurin series is $\sum_n a_n z^n$ and $\sum_n |a_n|$ converges, then $\sum_n a_n = S$ converges and 
Abel's theorem says $f(x) \to S$ as $x \to 1-$.  Of course $f(r_n) = 0$, so all we have to do is ensure that  $f(s_n)$ does not go to $0$ for some other sequence $s_n \to 1-$. 
Take $t_n = 1 - b^{n}$, $r_n = t_{2n}$ and $s_n = t_{2n+1}$, where $0 < b < 1/2$.  Note that if $m < n$, $(1-t_n)/(1-t_m) = b^{n-m}$ and
$$ \dfrac{t_n - t_m}{1-t_n t_m} = 1 - \dfrac{(2-b^m) b^{n-m}}{1 + b^{n-m} - b^{n}} > 1 - 2 b^{n-m}$$
Thus we get 
$$|f(s_n)| > \prod_{j=1}^\infty (1 - 2 b^{1+2j})^2 > 0$$
