# Is there function that can be expanded as infinite power series with bounded positive coefficients？

Is there a rational function $$F$$ which may be expanded as power series with coefficients of unperiodical positive integers in such a form:

$$F(x)=\sum_0^{\infty}a_i x^i,\qquad a_i\in \mathcal{N} \cup \{0\},\quad\exists M\forall i,\space a_i \leq M.$$

Any unclear meaning may be clarified by comment and discussion.

Second question: Let $$F$$ be a function which may be expanded as power series with coefficients of unperiodical positive integers as above. Prove that $$F$$ is transcendental (I guess a function with expansion of power series with coefficients of bounded unperiodical positive integers has to be a rational function or transcendental).

• what does unperiodical mean exactly? Eventually unperiodical, or something else? – user140765 Jun 1 at 14:33
• Don't the Taylor coefficients of a rational function always satisfy a linear recurrence relation? That together with boundedness should give you eventual periodicity. – Andreas Blass Jun 1 at 14:42
• The key ingredient is Kronecker's theorem: if $P \in {\bf Z}[X]$ is monic and all complex zeros satisfy $|z|\leq 1$ then those zeros are roots of unity. Let $P$ be the denominator of $F$ and soon conclude that the power-series coefficients are periodic. – Noam D. Elkies Jun 1 at 16:42
• The question specified integral (and positive) coefficients -- otherwise there are easy counterexamples like $1/(2-x)$. Yes, it must be proved that this implies that the denominator has constant coefficient $\pm 1$ (not "monic" as I mistakenly wrote above: one must substitute $1/x$ for $x$ to make that work). – Noam D. Elkies Jun 1 at 23:34
• I think it is worth recalling a theorem of Christol, Kamae, Mendès-France and Rauzy (“Suites algébriques, automates et substitutions”, Bull. SMF 108 (1980), 401–419): a formal series $f = \sum a_i x^i \in \mathbb{F}_q[[x]]$ with coefficients in a finite field (with $q=p^r$ elements) is algebraic if and only if its coefficient $a_i$ can be computed by a finite automaton from the base $p$ expansion of $i$. – Gro-Tsen Jun 2 at 10:04