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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.

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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 $f, \sum_{n=1}^{\infty}|a_n| = \infty.$

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

$$|f(s_n)| > \prod_{j=1}^\infty (1 – 2 b^{1+2j})^2 > 0$$

From the answer of @zhw. and the comments of @Robert Israel I think we can provide an explicit example as follows:

$$z \mapsto e^{-\frac{1+z}{1-z}}$$ is holomorphic on the unit disk and bounded in absolute value by $1$ ( @zhw. answer)

From the comment of @Robert Israel we get the explicit series expansion

$$e^{-\frac{1+z}{1-z}} = e^{ -1 -\frac{2z}{1-z}} = e^{-1} \cdot \sum_{n\ge 0} L^{(-1)}_n(2)\, z^n$$

where $L^{(\alpha)}_n(x)$ are the Laguerre polynomials.

We have asymptotics for the Laguerre polynomials. Some numerical testings suggest the following explicit bound:

$$ \frac{L^{(-1)}_n(2)}{e} – \frac{2^{1/4} }{\sqrt{\pi}} \frac{\cos(\sqrt{8n} + \frac{\pi}{4}) }{n^{3/4}} = \theta_n\cdot n^{-5/4}$$

with $|\theta_n| \le \frac{1}{7.2}$ ($n \ge 1$).

We conclude that for all $z$ in $\mathbb{C}$, $|z| <1$ we have

$$\left|\sum_{n\ge 1} \frac{\cos(\sqrt{8n} + \frac{\pi}{4}) }{n^{3/4}}\, z^n\right | < \frac{\sqrt{\pi}}{2^{1/4}}\cdot (\, 1/e + 1 + \frac{5}{36} \zeta(5/4)\,) = 2.9899..<3$$

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