Is there a result on the behaviour of power series with positive integer coefficients on their boundary? I have a power series whose coefficients are all positive integers and whose radius of convergence $r$ is $<1$ and I wish to prove that it has a pole at $r$, or at least an infinite radial limit. Is there a general result that could help in this situation or should I look for an ad-hoc proof?
 A: I can give you the following proof of what is an extension of Abel's Theorem.

THM Suppose $\alpha_n$ is a sequence of real numbers such that $\sum_{n\geqslant 0}\alpha_n$ diverges to $+\infty$, and such that its powerseries converges for $|x|<1$. Then $$\lim_{x\to 1^{-}}\sum_{n\geqslant 0}\alpha_nx^n=+\infty$$

P Let $M>0$ be given. Note that $$\frac{1}{1-x}f(x)=\sum_{n\geqslant 0}\sum_{k=0}^n\alpha_k x^n$$
By hypothesis this is true for $|x|<1$. Also, there exists $N>0$ such that $n>N$ implies $\sum_{k=1}^n\alpha_k>M$. Then we have that 
$$\displaylines{
  \frac{1}{{1 - x}}f(x) = \sum\limits_{n \geqslant 0} {\sum\limits_{k = 0}^n {{\alpha _k}} } {x^n} \cr 
   = \sum\limits_{n = 0}^N {\sum\limits_{k = 0}^n {{\alpha _k}} {x^n}}  + \sum\limits_{n > N} {\sum\limits_{k = 0}^n {{\alpha _k}} } {x^n} \cr 
   > \sum\limits_{n = 0}^N {\sum\limits_{k = 0}^n {{\alpha _k}} {x^n}}  + M\sum\limits_{n > N} {{x^n}}  \cr 
   > M{x^{N + 1}}\frac{1}{{1 - x}} \cr} $$
It follows that $f(x) > M{x^{N + 1}}$ so $$\mathop {\lim \inf }\limits_{x \to {1^ - }} f(x) \geqslant M$$ for each $M>0$, whence it must be the case 
$$\mathop {\lim }\limits_{x \to {1^ - }} f(x) =  + \infty $$
ADD Note the last inequality follows from the fact we can assume  the partial sums are all positive. 
A: You are looking for Pringsheim's theorem. I'm citing from Wikipedia's entry on Alfred Pringsheim: "One of Pringsheim's theorems, according to Hadamard [1] earlier proved by E. Borel, states [2] that a power series with positive coefficients and radius of convergence equal to 1 has necessarily a singularity at the point 1.". Of course you also get the same statement when the radius of convergence is $r$ by scaling the coefficients.
EDIT: Precisely if $f$ has positive coefficients and radius of convergence $r$ then $g(z) = f(z r)$ has radius of convergence $1$ and positive coefficients, therefore by Pringsheim's theorem $g$ has a singularity at $1$, hence $f$ has a singularity at $r$.
