Convergence between $\sum |a_n|$ and $\sum a_n z^n$ for all $z\in\mathbb{T}$ Let $\{a_n\}_{n=0}^{\infty}$ be a sequence of complex numbers. $\mathbb{T}$ is the unit circle over complex plane. It is obvious that if $\sum |a_n|$ converges, then $\sum a_n z^n$ converges for each $z\in \mathbb{T}$ fixed. Is the converse also true?
If the answer is “no”, what will happen if we strength the condition by assuming $\sum a_n z^n$ converges pointwise and defined a continuous function on $\mathbb{T}$?

Here are some observations.
1.Suppose $\sum a_n z^n$ converges pointwise on $\mathbb{T}$. Denote the limit function by $f$, then by Baire category theorem, $f$ is continuous except a set of first category.
2.I think this question may have some relations with the following question: find a function in disc algebra $A(\mathbb{D})$, and there is no analytic continuation of $f$.
 A: See also this posting in MO by me: Ordinary generating function for Mobius.
The answer to your question is  no  even in the case of pointwise convergence and continuous on $\mathbb{T}$.
We modify the ordinary generating function for Mobius function slightly. Let $$
f(z)=\sum_{n=1}^{\infty} \frac{\mu(n)}n z^n.
$$
The radius of convergence of the series is $1$, also the following argument shows the convergence on $\mathbb{T}$. Let $A_{\theta}(t)=\sum_{n\leq t} \mu(n)e^{2\pi i n \theta}$. By partial summation,
$$
\begin{align}
\sum_{n\leq x} \frac{\mu(n)}n e^{2\pi i n \theta} &=\int_{1-}^x \frac1t dA_{\theta}(t)=\frac{A_{\theta}(t)}t \Bigg\vert_{1-}^x +\int_1^x \frac{A_{\theta}(t)}{t^2}dt
\end{align}
$$
By Davenport's theorem, for any fixed $h>10$ and uniformly for $\theta\in\mathbb{R}$,
$$
A_{\theta}(t) = O\left( \frac t{\log^h t}\right).
$$
The implied constant in the big-oh is absolute. Taking $x\rightarrow\infty$,
$$
\sum_{n=1}^{\infty}\frac{\mu(n)}n e^{2\pi i n \theta} = \int_1^{\infty} \frac{A_{\theta}(t)}{t^2}dt,
$$
the integral converges absolutely.
Fix $\theta_0\in\mathbb{R}$. By the uniformity in Davenport's theorem and the Dominated Convergence Theorem, we have
$$
\lim_{\theta\rightarrow\theta_0} \sum_{n=1}^{\infty} \frac{\mu(n)}n e^{2\pi i n \theta}=\lim_{\theta\rightarrow\theta_0}\int_1^{\infty} \frac{A_{\theta}(t)}{t^2}dt=\int_1^{\infty} \frac{A_{\theta_0}(t)}{t^2}dt = \sum_{n=1}^{\infty} \frac{\mu(n)}n e^{2\pi i n \theta_0}.
$$
Hence, we have the convergence of $f(z)$ for any $z\in \mathbb{T}$ and the pointwise limit is continuous on $\mathbb{T}$.
However, this series is not absolutely convergent. To see this, consider
$$
\sum_{n=1}^{\infty} \frac{|\mu(n)|}n = \infty. 
$$
