Still thinking about the interesting question!

Not an answer, but too big for a comment.

To show what I meant in my comment to Daniel Litt's answer about the difference between uniform absolute convergence and ordinary uniform convergence:

I think (but it was a while ago when I thought about it) that there exist $u_0, u_1, \ldots$ such that $\sum_{n=0}^\infty u_n z^n$ converges uniformly on the set $\{ z \in \mathbb{C} : |z| \leq 1 \}$, but not uniformly absolutely.

Thus $\sum_{n=0}^\infty |u_n| = +\infty$, but also $$ \forall \\, \epsilon>0, \quad \exists \\, N(\epsilon)>0 \quad \text{such that}: \qquad \forall \\, |z| \leq 1, \quad \forall \\, m \geq n > N(\epsilon), \qquad \left| \sum_{k=n}^{m} u_k z^k \right| < \epsilon. $$ This function $f(z) = \sum_{n=0}^\infty u_n z^n$ satisfies $$ f \in A(D) \setminus W_+(D), $$ which shows in particular that the disc algebra $A(D)$, consisting of functions continuous on the closed unit disc and analytic on the open unit disc, is strictly larger than the Wiener algebra $W_+(D)$ of power series absolutely convergent on the closed unit disc.

So the problem is going to be pretty hard because we can't use absolute convergence (unless I'm being stupid or there is a clever trick exploiting the special structure of the exponential functions).

The easiest "solution" is just to ignore it (i.e. assume absolute convergence!) This gives a slightly different, but still interesting, question.

Still thinking about the interesting question!

Not an answer, but too big for a comment.

To show what I meant in my comment to Daniel Litt's answer about the difference between uniform absolute convergence and ordinary uniform convergence:

I think (but it was a while ago when I thought about it) that there exist $u_0, u_1, \ldots$ such that $\sum_{n=0}^\infty u_n z^n$ converges uniformly on the set $\{ z \in \mathbb{C} : |z| \leq 1 \}$, but not uniformly absolutely.

Thus $\sum_{n=0}^\infty |u_n| = +\infty$, but also $$ \forall \, \epsilon>0, \quad \exists \, N(\epsilon)>0 \quad \text{such that}: \quad \forall \, |z| \leq 1, \quad \forall \, m \geq n > N(\epsilon), \quad \left| \sum_{k=n}^{m} u_k z^k \right| < \epsilon. $$ This function $f(z) = \sum_{n=0}^\infty u_n z^n$ satisfies $$ f \in A(D) \setminus W_+(D), $$ which shows in particular that the disc algebra $A(D)$, consisting of functions continuous on the closed unit disc and analytic on the open unit disc, is strictly larger than the Wiener algebra $W_+(D)$ of power series absolutely convergent on the closed unit disc.

So the problem is going to be pretty hard because we can't use absolute convergence (unless I'm being stupid or there is a clever trick exploiting the special structure of the exponential functions).

The easiest "solution" is just to ignore it (i.e. assume absolute convergence!) This gives a slightly different, but still interesting, question.

Still thinking about the interesting question!

Not an answer, but too big for a comment.

To show what I meant in my comment to Daniel Litt's answer about the difference between uniform absolute convergence and ordinary uniform convergence:

I think (but it was a while ago when I thought about it) that there exist $u_0, u_1, \ldots$ such that $\sum_{n=0}^\infty u_n z^n$ converges uniformly for all $|z| \leq 1$, but not uniformly absolutely, i.e. $\sum_n |u_n| = +\infty$, but also $$ \forall \\, \epsilon>0, \quad \exists \\, N(\epsilon)>0 \quad \text{such that}: \qquad \forall \\, |z| \leq 1, \quad \forall \\, m \geq n > N(\epsilon), \qquad \left| \sum_{k=n}^{m} u_k z^k \right| < \epsilon. $$ This function $f(z) = \sum_{n=0}^\infty u_n z^n$ satisfies $$ f \in A(D) \setminus W_+(D), $$ which shows in particular that the disc algebra $A(D)$, of continuous functions on the closed unit disc which are analytic on the open unit disc, is strictly larger than the Wiener algebra $W_+(D)$ of power series absolutely convergent on the closed unit disc.

So the problem is going to be pretty hard because we can't use absolute convergence (unless I'm being stupid or there is a clever trick exploiting the special structure of the exponential functions).

The easiest "solution" is just to ignore it (i.e. assume absolute convergence!) This gives a slightly different, but still interesting, question.

Still thinking about the interesting question!

Not an answer, but too big for a comment.

To show what I meant in my comment to Daniel Litt's answer about the difference between uniform absolute convergence and ordinary uniform convergence:

I think (but it was a while ago when I thought about it) that there exist $u_0, u_1, \ldots$ such that $\sum_{n=0}^\infty u_n z^n$ converges uniformly on the set $\{ z \in \mathbb{C} : |z| \leq 1 \}$, but not uniformly absolutely.

Thus $\sum_{n=0}^\infty |u_n| = +\infty$, but also $$ \forall \\, \epsilon>0, \quad \exists \\, N(\epsilon)>0 \quad \text{such that}: \qquad \forall \\, |z| \leq 1, \quad \forall \\, m \geq n > N(\epsilon), \qquad \left| \sum_{k=n}^{m} u_k z^k \right| < \epsilon. $$ This function $f(z) = \sum_{n=0}^\infty u_n z^n$ satisfies $$ f \in A(D) \setminus W_+(D), $$ which shows in particular that the disc algebra $A(D)$, consisting of functions continuous on the closed unit disc and analytic on the open unit disc, is strictly larger than the Wiener algebra $W_+(D)$ of power series absolutely convergent on the closed unit disc.

So the problem is going to be pretty hard because we can't use absolute convergence (unless I'm being stupid or there is a clever trick exploiting the special structure of the exponential functions).

The easiest "solution" is just to ignore it (i.e. assume absolute convergence!) This gives a slightly different, but still interesting, question.