About convergence of complex power series on the circle ##|z - z_0|=r##

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cianfa72
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Does convergence of power series on the circle ##|z - z_0|=r## imply absolute convergence there ?
Whether a complex power series ##\sum_{k=0}^\infty c_k (z - z_0)^k## converges at a point ##\tilde z \in \mathbb C## then it converges absolutely in the open disk ##|z−z_0|<|\tilde z−z_0|=r##.

Assume now a power series convergent on the circle ##|z−z_0|=r##, does it imply absolute convergence on the closed disk ##|z−z_0|≤r## ?

Searching on net, I found a negative answer as in this link from MSE: Sierpinski (1916) produced an example where the power series converges everywhere on the unit circle, yet the function ##f(z)## it converges to isn't bounded on the closed disk ##|z|≤1##, hence isn't continuous on it and certainly does not converge absolutely on the circle ##|z|=1##.

In the interior of the unit disk the function ##f(z)## is continuous, for absolute convergence there imply uniform convergence that in turn imply continuity. It follows that ##f(z)## is bounded inside the unit disk, hence it must be unbounded on some point/points on the circle ##|z|=1##.

Now my question is: how is actually possible that this power series converges also on this point/points on the circle ##|z|=1## as assumed by hypothesis ?
 
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