Power series with funny behavior at the boundary Consider a power series
$$
\sum_{n=0}^{\infty}a_nz^n
$$
where $a_n$ and $z$ are complex numbers. There is radius $R$ of convergence. Let us assume that is a positive real number. It is well known that for $|z|<R$ the series converges absolutely; for $|z|>R$ it does not converge.
On the other hand, when $|z|=R$, the series can have very different behaviors. This has been discussed in many posts, e.g. 
Behaviour of power series on their circle of convergence , Seeking a Geometric Proof of a Generalized Alternating Series' Convergence .
I am looking for some relatively easy explicit examples of power series with funny behavior at the boundary. The only I know is
$$
\sum_n\frac{1}{n}z^n
$$
or small variations, e.g. replacing $z$ with $z^k$.
thanks
 A: The series $$f(z) = \sum_{n=1}^\infty \dfrac{z^{2^n}}{n}$$
converges almost everywhere on the unit circle by Carleson's theorem (it is the Fourier series of an $L^2$ function).  However, it diverges on a dense set, including all the $2^k$'th roots of unity: 
in fact at each of those points the real parts of the partial sums $S_k = \sum_{n=1}^k z^{2^n}/n$ are unbounded above.  
Now for each positive integer
$N$, the set $U_N$ of points $z$ such that $\text{Re}(S_k) > N$ for some $k$
is open and dense in the unit circle.  The intersection $G$ of the $U_N$ is a dense 
$G_\delta$ by the Baire category theorem, and the series diverges at every point of $G$.
Putting these facts together, we find that the set of points of the unit circle where the series diverges is negligible in the sense of Lebesgue measure but the set where it converges is negligible in the sense of Baire category.  I'd call that funny...
A: Here is a funny result. $\newcommand{\ve}{\varepsilon}$ Consider the random series 
$$ \sum_{n\geq 1}\frac{\ve_n}{n} z^n, $$
where the signs  $\ve_n=\pm 1$ are chosen  randomly and independently with ${\rm Prob}\;(\ve_n=\pm 1)=\frac{1}{2}$, $\forall n$.
Then, almost surely, the above series cannot be continued across  the circle of convergence $|z|=1$; see Chapter 4 of Kahane's book on random series of functions. 
On the other hand, Kolmogorov's three-series theorem   implies that for any fixed $z$ with $|z|=1$ the  above series converges for almost any choice of random signs $\ve_n$.
Put these two facts together to deduce that, with probability  $1$,  the above series  cannot be  extended across $|z|=1$, yet it converges for any $z$ of the form $z=\exp( 2\pi \sqrt{-1} r)$, $r\in \mathbb{Q}$. 
