$B\subset \mathbb{N}\bigcup \{0\}$ 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?



Randomness about coefficients of series
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](http://en.wikipedia.org/wiki/Infinite-dimensional_Lebesgue_measure)).


