$B\subset \mathbb{N}\bigcup \{0\}$ is finite and not empty, infinite series:$$f(x)=\sum_{i=1}^{\infty}a_i x^i,a_i \in B$$ Now $f(x)$ is rational or has a natural boundary.
Now,the question :if $f(x)$ has a natural boundary, is $$\lim_{n\rightarrow \infty}K(a_1a_2\cdots a_i \cdots a_n)\rightarrow \infty$$ $K(a_1a_2\cdots a_i \cdots a_n)$ is Kolmogorov complexity. Or, the sequence $a_1a_2\cdots a_i \cdots $ is random?
If $B$ is infinite,when is $a_1a_2\cdots a_i \cdots $ random?
This question is not well formed.
First, I assume by $K(a_1 a_2 \ldots)$ you mean $K(\langle a_1, a_2, \ldots \rangle)$. As the size of the sequence increases (regardless of the choice of $a_i$), the complexity must increase to infinity.
On the other hand, there is nothing computationally complex about the coefficients of certain naturally bounded series. For example, http://planetmath.org/naturalboundary says that $\sum_{k=0}^\infty z^{k!}$ has a natural boundary. Clearly the coefficients $a_i$ are computable.
Last, it is not very well-defined what it means for a sequence $a_1,a_2,\ldots \in \mathbb{N}^\mathbb{N}$ to be random. In particular Martin-Löf randomness requires a locally finite measure to be well-defined (and usually one uses a probability measure). I don't think there is a natural choice of measure on $\mathbb{N}^\mathbb{N}$ (for the same reason there is no infinite dimensional Lebesgue measure).
