# Mode of convergence of a power series

I am looking for a power series $\displaystyle f(z) = \sum_{n=0}^{+\infty} a_n z^n$ that converge uniformly on $\mathcal{D} = \Big\{ z \in \mathbb{C} \ / \ \vert z \vert \leq 1 \Big\}$ but not normally. Of course, a proof that this is impossible would be even better. It seems close to this question, but it's not quite the same.

It is well-known that normal convergence implies uniform convergence, and that the converse is false but I haven't found yet a counterexample in the form of a power series on $\mathcal{D}$.

In other words, I would like a complex sequence $(a_n)_{n \in \mathbb{N}}$ satisfying the three conditions :

• $\displaystyle\sum_{n=0}^{+\infty} a_n z^n$ converges for all $z$ such that $\vert z \vert \leq 1$
• $\displaystyle \sup_{\vert z \vert \leq 1} \left\vert \sum_{n=N}^{+\infty} a_n z^n \right\vert \longrightarrow 0$ when $N\to+\infty$
• $\displaystyle\sum_{n=0}^{+\infty} \vert a_n \vert = +\infty$
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I had to look up "normal convergence": eom.springer.de/N/n067430.htm – Yemon Choi Jan 25 '10 at 22:31
Me too. Fortunately, as indicated at the end of the question, in this case it is the same as absolute convergence. (More generally it is the stronger condition that the hypotheses of the Weierstrass M-test are satisfied.) – Jonas Meyer Jan 26 '10 at 5:30