TITLE: Fraction field of $\mathbb Z_p[[X]]$
QUESTION [5 upvotes]: We know that the fraction field $F:=\operatorname{Frac}(\mathbb Z_p[[X]])$ is strictly contained in the field of Laurent power series $\mathbb Q_p((X))$, thanks to  this result of Gilmer. So my question is:

Is it possible to describe explicitly the elements of $F$?

Some similar questions have been already asked here or on Mathoverflow. Maybe the most relevant is this one regarding the explicit computation of the fraction field of $\mathbb Z[[X]]$. Someone suggests in the comments of the linked question that the problem with $\mathbb Z_p$ (instead of $\mathbb Z$) should be easier.
Some general necessary conditions are given here when the coefficients of the power series lie in any domain, but I'd like to find some sufficient conditions in the particular case of $\mathbb Z_p$.
Many thanks in advance

REPLY [4 votes]: Say you have a power series $\sum_k a_kX^k \in \mathbb Z_p[[X]]$.
If it's nonzero, you can write it as $X^np^m\sum_k b_kX^k$ with $b_0 \notin (p)$.
In particular, as $\mathbb Z_p$ is local, $b_0$ is invertible, and so $\sum_kb_k X^k$ is also invertible : you only have to invert $X^np^k$
In particular, $\operatorname{Frac}(\mathbb Z_p[[X]]) = \mathbb Z_p[[X]] [X^{-1}, p^{-1}]$.
So an element $f\in \mathbb Q_p((X))$ is in $\operatorname{Frac}(\mathbb Z_p[[X]])$ if and only if the $p^n$ in the denominators are bounded
(the above description shows the "only if" bit, and for "if" : if they are bounded, multiplying by $p^k$ for $k$ large enough makes you land in $\mathbb Z_p((X))$)
As YCor points out in the comments of the MO question about $\mathbb Z[[X]]$, the question is probably easier in local rings more generally, although here I have in fact used that the maximal ideal was principal (so this works over discrete valuation rings)