Complex power series converging at only one point on the unit circle The complex power series
$$\lambda(z) = \sum_{n=0}^\infty \frac{z^n}{n}$$
Has radius of convergence $R=1$, and converges for all complex $z$ on the unit circle except for $z=1$. Does anyone know of a complex power series
$$\sum_{n=0}^\infty a_n z^n$$
for which the opposite is true - so that it also has radius of convergence $R=1$, but diverges for all complex $z$ on the unit circle except $z=1$? Or is there a reason why no such power series could exist?
 A: The following content abbreviates pp. 69-71 of Edmund Landau's book Darstellung und Begründung Einiger Neuerer Ergebnisse der Funktionentheorie, with some symbols changed to my liking.

Theorem (Lusin) There exists $f(z) = \sum_{n=0}^\infty a_n z^n$ such that $a_n \to 0$, which diverges everywhere on the unit circle

Proof: For each $m>1$, let $$g_m(z) = 1+z+\cdots+z^{m-1} = \frac{1-z^m}{1-z}$$
For $\zeta = e^{i\varphi}\neq1$, we have
$$|g_m(\zeta)| = \left| \frac{\sin\frac{m\varphi}{2}}{\sin\frac{\varphi}{2}}\right|$$
On the arc $-\frac{\pi}{m}\leq\varphi\leq\frac{\pi}{m}$, excluding $\varphi=0$, we have
$$|g_m(\zeta)| \geq \frac{\frac{2}{\pi}\left|\frac{m\varphi}{2}\right|}{\left|\frac{\varphi}{2}\right|} = \frac{2m}{\pi}$$
which holds because $\left|\frac{2x}{\pi}\right| \leq |\sin x| \leq |x|$ for $-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}$ (Of course, this estimate also holds for $\varphi = 0$). One can similarily show that
$$\max_{0\leq k < m}\left|g_m\left(e^{-\frac{2k\pi i}{m}}\zeta\right)\right|\geq \frac{2m}{\pi}$$
For each $m>0$, consider the polynomial given by
$$h_m(z) = \sum_{k=0}^{m-1} z^{mk} g_m\left(e^{-\frac{2k\pi i}{m}}z\right)$$
$h_m(z)$ has degree $(m-1)(m+1)$, and by construction each coefficient of $h_m(z)$ is an $m$-th root of unity, hence has modulus 1. Finally, consider the power series given by
$$f(z) = \sum_{n=0}^\infty a_n z^n = \sum_{m=1}^\infty \frac{h_m(z)}{\sqrt{m}} z^{1^2 + 2^2 + \cdots (m-1)^2} $$
One can easily check that $|a_n|$ follows the sequence $1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}, \cdots,\frac{1}{\sqrt{m}}\,(m^2\,\text{times}),\cdots$ hence $a_n \to 0$. In particular it is obvious that the radius of convergence of $f$ must be at least 1. We shall now show that $f$ diverges at every point on the unit circle, so that the radius of convergence must be exactly 1. Indeed, were $f(\zeta)$ to converge for some $|\zeta|=1$, then we would have
$$\lim_{m\to\infty} \frac{1}{\sqrt{m}}\left|\zeta^{1^2+\cdots+(m-1)^2}\right|\max_{0\leq k < m} \left|\zeta^{mk} g_m\left(e^{-\frac{2k\pi i}{m}}\zeta\right)\right| = 0$$
however
$$\frac{1}{\sqrt{m}}\max_{0\leq k < m} \left|g_m\left(e^{-\frac{2k\pi i}{m}}\zeta\right)\right|\geq \frac{2\sqrt{m}}{\pi}$$
so this cannot happen. It follows that $f$ diverges at all points on the unit circle.

Corollary (Sierpiński) There exists $g(z) = \sum_{n=0}^\infty b_n z^n$ which diverges at all points on the unit circle except $z=1$.

Proof: Suppose $f(z) = \sum_{n=0}^\infty a_n z^n$ as in Lusin's example. Consider
$$g(z) = a_0 - a_0 z + a_1 z^2 - a_1 z^3 + a_2 z^4 - a_2 z^5 \cdots$$
This series converges to $0$ at $z=1$ since $a_n \to 0$. But if $g$ were to converge at any $\zeta \neq 1$ on the unit circle, we would have that
$$a_0 (1-\zeta) + a_1\zeta^2(1-\zeta) + a_2\zeta^4(1-\zeta)+\cdots$$
converges, hence so does
$$f(\zeta^2) = a_0 + a_1 \zeta^2 + a_2 \zeta^4 + \cdots$$
but $|\zeta^2| = 1$ so this cannot happen. So $g$ is a power series with radius of convergence $1$, which converges at $z=1$ but diverges elsewhere on the unit circle.
Using a more involved construction based on the same ideas, we also have the following two theorems

Theorem 1 (Herzog and Piranian) For any $F_\sigma$ subset $M$ of the unit circle $C$, there exists a power series which converges
everywhere in $M$ and diverges everywehre in $C\setminus M$


Theorem 2 If $M$ is a subset of the unit circle $C$, such that some power series converges on $M$ and diverges on $C\setminus M$,
then $M$ is a $F_{\sigma\delta}$ set.

Proofs of these two statements can be found in Herzog, Fritz; Piranian, George (1949). Sets of convergence of Taylor series I. Duke Mathematical Journal, 16(3), 529–534. As mentioned in the corresponding discussion on MO, the converse of Theorem 2 is not true:

Theorem (Lukašenko) There is a $G_\delta$ subset of the unit circle which is not a set of convergence of some power series.

Further, the characterization of precisely which subsets of the unit circle can be sets of convergence remains an open problem today.
Edit: According to the comments of the MO answer, the actual reference for the result on $G_\delta$ sets originates from Lukašenko S. Ju., Sets of divergence and nonsummability for trigonometric series, Vestnik Moskov. Univ. Ser. I Mat. Mekh. 1978, no. 2, 65–70. Rather, Körner wrote a survey paper which includes this result.
