TITLE: Is there a local-global principle for integral Laurent series ? 
QUESTION [5 upvotes]: Motivation: A real number is rational iff its decimal expansion is periodic (by "periodic" I mean periodic after some steps). Similar, a p-adic number is rational iff its p-adic expansion is periodic. However, this isn't true in general for rational functions as the example 
$$\frac{1}{(1-X)^2}= \sum_{n=0}^\infty (n+1)X^n \in \mathbb{Q}(X)\subseteq \mathbb{Q}((X))$$
shows. But an inspection of the proof in the p-adic case (cf. Hasse: Number Theory, Chap. 9) shows: 

If $F$ is a finite field, then $f \in F((X))$ is rational, i.e. $f \in F(X)$, iff its Laurent series 
  is periodic.

Now let $f=g/h$ with polynomials $g,h \in \mathbb{Z}[X]$ such that the Laurent series of $f$ has  integer coefficients. Reduction modulo p yields $\bar{f}=\bar{g}/\bar{h} \in\mathbb{F}_p(X)$. Hence the coefficients of $f$ are periodic modulo p. I wonder  whether the converse is true:

Question 1: Let $f$ be a Laurent series with integer coefficients. Is $f \in Quot(\mathbb{Z}[X])$ iff the coefficients of $f$ are periodic for all primes (with period depending on the prime) ?

Suppose that the coefficients of $f$ are periodic modulo p. Hence the reduction modulo p 
is rational, i.e. $\bar{f} \in \mathbb{F}_p(X)$.  Therefore an equivalent formulation of the question is: 

Question 2: Let $f$ be a Laurent series with integer coefficients. 
  Is $f$ rational, i.e. $f \in Quot(\mathbb{Z}[X])$ iff the reduction modulo p is rational for all primes ? 


Edit: As shown by Felipe, the answer is in general negative. Since I'm mostly interested in convergent power series (with integer coefficients), I would like to ask in addition, if there are also counterexamples in this case ?

REPLY [9 votes]: No $\sum n! x^n$ is a counterexample. I think it's true for algebraic functions, though.