TITLE: Computation of fraction field of formal series over the integers
QUESTION [21 upvotes]: What is the fraction field $K$ of the domain $\mathbb Z[[X]]$?
It is strictly smaller than the  field of Laurent series $L=\operatorname {Frac}\mathbb Q[[X]]$, since $\sum_{i\geq 0}\frac {X^i}{i!}\in L\setminus K$.
Indeed a Laurent series $f(X)=\sum_{i\in \mathbb Z} q_iX^i \in L \; (q_i=0 \operatorname {for} i\lt\lt 0)$ must have as its coefficients rational numbers whose denominators involve only finitely many primes (which  depend only on $f(0)$) in order to belong to $K$.
But is that necessary condition sufficient?
Edit
No, that necessary condition is not sufficient.
For example if $p$ is a prime the series $\sum_{i\geq 0}\frac {X^i}{p^{i^2}}$ does not belong to $K=\operatorname {Frac}\mathbb Z[[X]]$, even though the denominators in its coefficients only involve the single prime $p$.
This follows from Elad Paran's Example 2.3(c) quoted by Arno in his very pertinent answer below.

REPLY [2 votes]: This is actually a remark on Arno Fehm's answer, but too long for a comment (the remark is due to P. Samuel). Let $p$ be a prime, and let us look at power series $u(X)=\sum_{n\geq 1}a_nX^n$ in $\mathbb{Q}((X))$ satisfying $u^2-pu+X=0$. This is equivalent to $a_1=p^{-1}$, $a_n=a_1a_{n-1}+\ldots +a_{n-1}a_1$. So the $a_i$ are uniquely determined, and of the form $\dfrac{b_i}{p^i} $ with $b_i\in\mathbb{Z}$. Still $u$ does not belong to $K$, because it is integral over $\mathbb{Z}[[X]]$ which is integrally closed. If it were in $K$, it would belong to $\mathbb{Z}[[X]]$, and this is clearly not the case.