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!

  • 2
    $\begingroup$ The partial sums converge uniformly on the full closed disc in this case so the limit is continuous on the full closed disc just by general topological results as asked $\endgroup$
    – Conrad
    Nov 30, 2023 at 14:03

1 Answer 1


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.

  • $\begingroup$ Thank you very much! In the end it boils down to Weierstrass M Test, doesn’t it? $\endgroup$ Nov 30, 2023 at 17:23
  • $\begingroup$ yes if you want to think it so - in general normal convergence (absolute and uniform on compacts) tends to boil down to that so when it holds it is much easier to prove than say (local) uniform convergence that is not absolute $\endgroup$
    – Conrad
    Nov 30, 2023 at 17:42
  • $\begingroup$ Is there any general theory of normal convergence in topology? $\endgroup$ Nov 30, 2023 at 18:49
  • $\begingroup$ not sure - usually this is a complex analysis notion but it may be generalized to other settings; the Cauchy integral representation makes it powerful since integrals behave well wr uniform convergence on compacts so it definitely would make sense in other situations where one has classes of fucntions defined by various local integral kernels $\endgroup$
    – Conrad
    Nov 30, 2023 at 19:16

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