Let $V$ be an arbitrary set of infinitely many positive integers, and let: $$\varsigma\_{V}\left(z\right)\overset{\textrm{def}}{=}\sum\_{v\in V}z^{v}$$ Let $T\_{V}$ denote the set of all $t\in\left[0,1\right)$ for which the limit:


$$c\_{V}\left(t\right)\overset{\textrm{def}}{=}\lim\_{x\uparrow1}\left(1-x\right)\varsigma\_{V}\left(e^{2\pi it}x\right)$$
exists. 


Suppose $0\in{T\_{V}}$ and $c\_{V}\left(0\right)\neq0$ (that is, $V$ has a positive, well-defined natural density). Are there then any $t\in\left(0,1\right)$ which *must* be in $T\_{V}$? More generally, what, if anything, can be said about $T\_{V}$ and/or $\left[0,1\right)\backslash T\_{V}$? (Ideally, all of $\mathbb{Q}$ (or all but some exceptional set) would have to be a subset of $T\_{V}$.) Conversely, what (if anything) does $\mathbb{Q}\subseteq T\_{V}$ then force to be true about $V$? In particular, if $\varsigma\_{V}\left(z\right)$ is a rational function, then $\mathbb{Q}\subseteq T\_{V}$ necessarily holds. Thus, another way of phrasing the question is: *does the set $\left\{ \varsigma\_{V}\left(z\right):T\_{V}\supseteq\mathbb{Q}\right\}$ contain functions which are* not *rational, and, if such functions do exist, what can be said about them and their associated $V$s?*



Existence of radial limits of products of certain power series and $1-x$
There are lots of different questions going on here, but here are a few observations.


First of all, if you change $V$ by any set with (upper) density zero, then $c\_V(t)$ doesn't change for any value of $t$. So for any $T\_V$ there are uncountably many possible choices of $V$, and so almost all of them will be transcendental functions. (For an explicit example where $T\_V = [0,1)$, let $V\_S$ consist of the set of all integers minus any subset $S$ consisting of powers of $2$. The set of possible $S$ is uncountable.)


Second, it's also relatively easy to construct $V$ such that $T\_V$ is missing infinitely many rational numbers. Let's start out with an example which is just missing $1/3$ and $2/3$. The idea is to take $V$ to consist of a string of numbers which are $1$ mod $3$, then a much longer string of numbers which is $2$ mod $3$, then a much much longer string which is $1$ mod $3$ and so on. That is, 


$$\varsigma(x) = \frac{x}{1-x^3} + x^{3a\_1} \frac{x^2 - x}{1-x^3} + x^{3 a\_2} \frac{x - x^2}{1 - x^3} + x^{3 a\_3} \frac{x^2 - x}{1-x^3} + \ldots $$


Note that


$$\varsigma(x) = \frac{x}{1-x^3} \left(1 + (x-1) x^{3a\_1} - (x-1) x^{3 a\_2} + (x-1) x^{3 a\_3} + \ldots \right)$$


Here we imagine that the $a\_n$ are growing really really fast, (say $x = n!^3$ or something). 
 Let $t = 1/3$ and let $\omega = e^{2 \pi i t}$.
Then $c\_V(1/3)$ exists if and only if


$$\lim\_{x \rightarrow 1^{-}} \left(1 + (\omega x-1) x^{3a\_1} - (\omega x-1) x^{3 a\_2} + (\omega x-1) x^{3 a\_3} + \ldots \right)$$


exists. The claim is that the value of this sum will vary between $1$ and $\omega$ as $x$ approaches $1$. The reason is that when $a\_n$ increases sufficiently fast, one can choose $x$ so that the first $n$ values of $x^{3 a\_n}$ are very close to one and the others decrease super-exponentially. For example, 
choose a value of $x$ such that


$$1 - \frac{1}{n^2 a\_n} < x < 1 - \frac{n}{a\_{n+1}} < 1 - \frac{n^{k}}{a\_{n+k}}.$$


Then for $m \le n$, 


$$x^{3 a\_m} \ge x^{3 a\_n} \ge \left(1 - \frac{1}{n^2 a\_n}\right)^{3 a\_n}
\sim 1 + O\left(\frac{1}{n^2}\right),$$


but for $m = n+k > n$, 


$$x^{3 a\_m} \le \left(1 - \frac{3 n^{k}}{a\_{n+k}}\right)^{3 a\_{n+k}}
\sim e^{-3 n^k}$$


which are thus negligible. Thus the value of the sum is approximately


$$1 + (\omega - 1) - (\omega -1) + (\omega -1) - \ldots $$


and so will vary between $1$ and $\omega$ and not converge.


On the other hand, it's not so hard to see that there will be no convergence issues for all *other* rational numbers except for $t = 1/3$ and $t = 2/3$. The point is that the $1-x$ term kills of any power series with terms that are so spread out.


But now consider:


$$\varsigma\_W(x):=\varsigma\_V(x) + \varsigma\_V(x^3) + \varsigma\_V(x^9) + \ldots $$


Since $V$ consisted only of terms $1$ and $2$ modulo $3$, there exists a corresponding set $W$. But now the analysis above shows that $T\_W$ avoids the set $a/3^n$ for every $(a,3) = 1$ and $n \ge 1$. The point is that these values are screwed up exactly by $\sigma\_V(x^{3^{n-1}})$. So $T\_W$ is missing infinitely many rational numbers.


Some preliminary analysis suggests one can construct functions missing all rational numbers except zero but the analysis could be a bit painful. (The idea is similar to the above, but to slowly increase the modulus one is considering from $2!$ to $3!$ to $4!$ etc.)


Other observations:


Certainly a $V$ chosen *randomly* will, almost always (in the appropriate sense) have $S \subset T\_V$ for any finite set $S$. I might even guess that for a random $V$ the set $T\_V$ contains all rational numbers, did you try to prove that? 


