Does a power series converging everywhere on its circle of convergence define a continuous function? Consider a complex power series $\sum a_n z^n \in \mathbb C[[z]]$ with radius of convergence $0\lt r\lt\infty$ and suppose that for every $w$ with $\mid w\mid =r$ the series $\sum a_n w^n $ converges .
We thus obtain a complex-valued function $f$ defined on the closed disk $\mid z\mid \leq r$ by the formula $f(z)=\sum a_n z^n$.  
My question: is $f$ continuous ?  
This is a naïve question which looks like it should be answered in any book on complex analysis.
But I checked quite a few books, among which the great treatises : Behnke-Sommer, Berenstein-Gay, Knopp, Krantz, Lang, Remmert, Rudin, Stein-Shakarchi, Titchmarsh, ... .
I couldn't find the answer, and yet  I feel confident  that it was known in the beginning of the twentieth century.  
Edit
Many thanks to Julien who has answered my question: Sierpinski proved (in 1916) that there exists such a power series  $\sum a_n z^n    $ with radius of convergence $r=1$ and associated function  $f(z)=\sum a_n z^n $  not  bounded on the closed unit disk and thus certainly not continuous. 
It is strange that not a single book on complex functions seems to ever have mentioned this example.   
On the negative side, I must confess that I don't understand Sierpinski's article at all!
  He airdrops a very complicated, weird-looking power series and proves that it has the required properties in a sequence of elementary but completely obscure computations.
I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing.
 A: Just to complete the previous answer, Sierpiński's example is mentioned (without details, though) in at least one book, namely An introduction to classical complex analysis by R. B. Burckel (Vol. 1, Chap. 3, p. 81).
A: I searched all over for an answer to this question back in my student days. I found the answer in a paper by Sierpinski, "Sur une série potentielle qui, étant convergente en tout point de son cercle de convergence,représente sur ce cercle une fonction discontinue ", which is featured in his collected works, see here, p282) and apparently was published in 1916.
It does confirm your expectation that this was known in the beginning of the twentieth century (I don't know whether it's the first proof or not, but from the paper it's clear that Sierpinski thought the result to be new).
EDIT: I just realized that not everybody speaks French ;-) so, to be clear: Sierpinski produces an example where the function converges everywhere on the unit circle but is discontinuous on the circle.
A: This answer is in response to final sentence, "I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing". In fact, it is easy to construct power series converging on the circle of convergence, but are unbounded. For example,
$$
f(z)=\sum_{n=1}^\infty\frac1{n^5(1+in^{-3}-z)}
$$
defines a function whose power series expansion has radius of convergence 1 and converges everywhere on the unit circle, but is unbounded in a neighbourhood of 1.
A method of constructing such functions is as an infinite sum
$$
f(z)=\sum_{n=1}^\infty f_n(z).
$$
Here, $f_n(z)$ are chosen to have a power series expansion converging everywhere on the closed unit ball. Let $f^{(r)}_n(z)$ denote the sum of the first $r$ terms in the power series expansion of $f_n$. We need to arrange it so that $f^{(r)}(z)\equiv\sum_nf_n^{(r)}(z)$ converges on the closed unit ball, and that $f(z)=\lim_{r\to\infty}f^{(r)}(z)$ holds. That is, we need to be able to commute the limit $r\to\infty$ with the summation over $n$. A sufficient condition to be able to do this is that $\sum_n\sup_r\lvert f^{(r)}_n(z)\rvert < \infty$, for all $\lvert z\rvert\le1$. That this allows us to commute the summation with the limit is just a special case of dominated convergence.
Next, to ensure that $f(z)$ is unbounded on the unit ball, we want to choose $f_n$ such that there exists $q_n$ in the closed unit ball with $f_n(q_n)$ large, and such that it does not get cancelled out in the summation, so that $f(q_n)$ is large and diverges as $n\to\infty$.
For example, choose positive reals $\delta_n,\epsilon_n$ tending to zero, and setting $a_n=1+i\epsilon_n$, and
$$
f_n(z)=\frac{\delta_n}{a_n-z}=\sum_{m=0}^\infty \delta_na_n^{-m-1}z^m.
$$
These are all well-defined as power series with radius of convergence greater than 1. Furthermore, the partial sums are
$$
f^{(r)}_n(z)=\delta_n\frac{1-(z/a_n)^r}{a_n-z},
$$
which are bounded by $2\delta_n/\lvert a_n-z\rvert$. As $a_n\to1$, this is bounded by a multiple of $\delta_n$ for each fixed $z\not=1$, so the dominated convergence condition is satisfied when $\sum_n\delta_n$ is finite. On the other hand, if $z=1$, then $\lvert a_n-z\rvert=\epsilon_n$, so the dominated convergence condition is satisfied everywhere whenever $\sum_n\delta_n/\epsilon_n$ is finite.
Next, $f_n(z)$ achieves its largest value on the unit ball at $q_n=a_n/\lvert a_n\rvert$, and its real part there is given by
$$
\Re f_n(q_n)=\frac{\delta_n}{\sqrt{1+\epsilon_n^2}(\sqrt{1+\epsilon_n^2}-1)}\ge\frac{2\delta_n}{\epsilon_n^2\sqrt{1+\epsilon_n^2}}.
$$
As $f_m(z)$ has positive real part for all $m$, this bound also holds for $f(q_n)$, and we get that $f$ is unbounded whenever $\delta_n/\epsilon_n^2\to\infty$. These conditions are satisfied by taking $\epsilon_n=n^{-3}$ and $\delta_n=n^{-5}$.
Alternatively, for an example closer to Sierpinski's, consider choosing a sequence $a_n\to1$ on the unit circle and positive reals $K_n$, and set
$$
f_n(z)=K_n2^{-n}\sum_{k=0}^{2^n-1}a_n^{2^n-1-k}z^k=2^{-n}K_n\frac{a_n^{2^n}-z^{2^n}}{a_n-z}.
$$
The partial sums of the power series expansion of $f_n(z)$ are bounded by $2^{1-n}K_n/\lvert a_n-z\rvert$, so the dominated convergence condition is satisfied for $z\not=1$ so long as $\sum_n2^{1-n}K_n$ is finite. Sierpinski chooses $a_n=(n^2-1+2ni)/(n^2+1)$ so that $a_n-1$ goes to zero at rate $1/n$. The dominated convergence condition is therefore satisfied whenever $\sum_n2^{-n}K_nn$ is finite.
Now, $f_n(z)$ is maximized at $z=a_n$ where $\lvert f_n(a_n)\rvert=K_n$. So,
$$
\lvert f(a_n)\rvert\ge K_n-\sum_{m\not=n}\frac{2^{1-m}K_m}{\lvert a_m-a_n\rvert}.
$$
As $a_m-a_n$ is bounded below by a multiple of $1/m^2$, the summation on the right is bounded whenever $\sum_m2^{-m}K_mm^2$ is finite, and $f(a_n)$ is unbounded if we also take $K_n$ going to infinity. Sierpinski takes $K_n=n^2$ here. Finally, in Sierpinski's example, he multiplies $f_n$ by $z^{2^n}$. This changes nothing, except to separate out the non-zero terms of the power series of $f_n(z)$, so that the power series of $f(z)$ can be written easily term by term.
