It is not hard to see that the field of fractions of $\mathbb{Z}[X]$ is the same as the field of fractions of $\mathbb{Q}[X]$; so I wonder if the same is true for the respective power series rings.
My attempt: Let $p(x) \in \mathbb{Z}[[X]]$ such that its constant term is nonzero. In $\mathbb{Q}[[X]]$, $p(x)$ is a unit; consequently, $p(x)/1$ is an element of the field of fraction of $\mathbb{Q}[[X]]$. On the other hand, $p(x)$ is not a unit of the integral domain $\mathbb{Z}[[X]]$. Then $p(x)/1$ is not an element of the field of fractions of $\mathbb{Z}[X]$. Am I right? Thanks in advance!
$Frac(\Bbb{Z}[[X]])$ doesn't contain $f(X)=\sum_{n=0}^\infty \frac{X^n}{n !} \in \Bbb{Q}[[X]]$ because there is no $g(X)\in a+X\Bbb{Z}[[X]]$ such that $f(X)g(X)= \sum_{m \ge 0} c_m X^m \in \Bbb{Z}[[X]]$ (for $p $ prime $p c_p \equiv \frac{a}{(p-1)!} \bmod p$)
Let $h(X) \in \Bbb{Z}[[X]]$ then $h(X) = b X^k (1+ \frac{X}{b} H(X))$ with $H(X) \in \Bbb{Z}[[X]]$ so that $$\frac{1}{h(X)} = \frac{X^{-k}}{b} (1+\sum_{l=1}^\infty \frac{(-X)^l H(X)^l}{b^l})$$ and hence $$Frac(\Bbb{Z}[[X]]) \subset \{ f(X) \in \Bbb{Q}((X)), \exists b,c,k \in \Bbb{Z}_{\ge 1}, c X^k f(bX) \in \Bbb{Z}[[X]]\}$$