Connection Between Convergence on Natural Boundary and Weierstrass Functions So, I was fooling around thinking about constructing functions on the unit disc $\mathbb{D}$, which cannot be extended to the boundary by Hadamard's Gap Theorem. At first I constructed the function 
$$f(z):=\sum_{k=1}^{\infty}{\frac {z^{k!}}{k^2} }$$
Since, I know that it will converge absolutely even on $\partial\mathbb D$. When I plotted the Real part of this function on the boundary, I noticed a very familiar Weierstrass-like function. (Which is not surprising to me after thinking about what this series actually is on the boundary)
My first question is: 1) Is the Real part of a function constructed similarly to this always a nowhere differentiable function of $\theta$ on the boundary of $\mathbb D$.
And my second question is: 2) It appears that 
$$g(z):=\sum_{k=1}^{\infty}{\frac {z^{k^2}}{k^2} }$$
also looks nowhere differentiable on the boundary by inspection on Mathematica. But, 
$$\lim_{k\to \infty}{\frac {(k+1)^2}{k^2}}=1$$
so, I don't believe Hadamard's Gap Theorem can be applied here. Is this function also not extendable to a holomorphic function on any larger domain as well?
(In general, if you could comment on the connection between non-differentiability on the boundary and extendability of the power series to larger domains at points on the boundary.) Thanks.
 A: This is not an answer to most of your question; please don't send money. In particular the question about $\sum z^{k^2}/k^2$ seems like it may be hard (except that it reminds me of someone's nowhere-differentiable function that turned out not to be nowhere differentiable; don't recall the details, sorry).
But I can comment on your last question. There exists a function in the unit disk that has the unit circle for natural boundary, but which is nonetheless infinitely differentiable on the closed disk!
Hmm. It's going to  be $$\sum c_kz^{2^k},$$where we may as well take $0<c_k\le 1$. So the radius of convergence is at least $1$; you're evidently aware that if the radius of convergence is $1$ then the unit circle is the natural boundary. So we need $$\sum_k c_k r^{2^k}=\infty\quad(r>1)$$to make the unit circle the natural boundary. To make it easier to see how this fits in with the other part of the argument we can rewrite this condition as $$\sum_k c_k 2^{\alpha2^k}=\infty\quad(\alpha>0).$$
To make the function $C^\infty$ on the closed disk we need $$\sum_kc_k2^{nk}<\infty\quad(n=1,2,\dots).$$So let $$c_k=2^{-k^2}.$$
A: You can take a series with convergence radius $1$, like
$$\sum_{n \ge 0} \frac{z^n}{e^{\sqrt{n}}}$$
and apply a mask to the terms ( keep only the powers of two indexes). This preserves the radius of convergence but introduces Hadamard gaps that makes the function un-extendable outside the open unit disk ( example from Rudin). However, this function, with whatever mask, is smooth on the closed unit disk. So, $C^{\infty}$ on the circle is not enough to extend analytically. However, analytic on any small region of the circle is enough the extend past that region of the circle. 
The second question is interesting indeed, no Hadamard gap, the squares mask on $\sum_{n\ge 1} \frac{z^n}{n}$. This one has a singular point only $1$, clearly $1$ stays singular since the terms are positive, not clear what other points are.
