Let $f(z) = \sum_{n=0}^{+\infty}c_nz^n$ be a complex power series with radius of convergence $R=1$. Suppose that the series of coefficients converges absolutely, i.e. $\sum_{n=0}^{+\infty}|c_n| < +\infty$. Is this conditon enough to prove that $f$ is continuous on the boundary of the disk of convergence?
I know that, by Abel's Theorem, the limit of $f$ at any point of this boundary is the power series evaluated at that point if we restric ourselves to a Stoltz sector. However, does the fact that this is true for every point of the boundary solve the problem of divergence with a tangential approach?
I would like to know wether this is true and, if that is the case, a sketch of the proof, or if there is some obvious counterexample that I am not considering as of now.
Any comment or answer is much appreciated and let me know if I can explain myself more clearly!
Best Answer
As requested, let's make my comment as an answer. Note that if we consider $f_N(z)=\sum_{n=0}^{N}c_nz^n$ we have by the triangle inequality $|f_N(z)-f(z)| \le \sum_{n \ge N+1}|c_n|$ whenever $|z| \le 1$, so the convergence of $\sum |c_n|$ implies that $f_N\to f$ uniformly on the closed unit disc which is a closed set, hence by general topology (and very easy to prove by the usual arguments) $f$ must be continuous on the closed unit disc too since $f_N$ are so.