Let us define the field of formal Laurent series over a field $k$ as $K=k((x_1))((x_2))...((x_n))$. The subring of formal Taylor series $R=k[[x_1,...x_n]]$ is embedded in this field. Let us call its field of fractions $K_0$. Does $K_0$ equal to $K$? I believe that the answer is no, but I can't prove it. If not, what is a necessary and sufficient condition for an element of $K$ to lie in $K_0$?
UPD: Here's one necessary (but I believe not sufficient) condition for an element of $K$ to lie in $K_0$. Let $a\in K$ when we have the series of two variables. Let us write it in the following standard form:
$$ a = \sum\limits_{n=-\infty}^{\infty}x^{-\varphi(n)}c_n(x)y^n, $$ where $c_n(x)$ is either a zero series or a series with a non-zero free term. If $c_n(x)=0$ then we suppose $k(n)=0$. Then the condition is the following:
If $a\in K_0$ then there exists $c\in\mathbb R$ such that for each $n$, $\frac{\varphi(n)}{n}<c$.
PROOF: Suppose $a\in K_0$. Then we can write it as $a=\frac{b}{c}$, $b,c\in R$. We denote $k[[x]]=S$. We write $c$ as
$$ c = y^k(r-y\alpha(y)) = ry^k(1-r^{-1}y\alpha(y)), $$ where $r\in S, \alpha(y)\in S[[y]]$. Then $$ c^{-1} = \sum\limits_{n=0}^{\infty}r^{-(n+1)}y^{n-k}\alpha^{n-k}(y) = \sum\limits_{n=-k}^{\infty}r^{-(n+k+1)}y^n\alpha^n(y). $$ We write $r=x^mr_0(x)$, where $r_0(x)\in k[[x]]^*$. And the last computation: $$ c^{-1} = \sum\limits_{n=-k}^{\infty}x^{-m(n+k+1)}r_0^{-(n+1)}(x)y^n\alpha^n(x,y). $$ From the equation above we see that $\varphi(n)\le m(n+k+1)$ as $\alpha(x,y)$ contains $x$ only in non-negative degrees. The series $b$ also contains $x$ only in non-negative degrees so after multiplication of $c^{-1}$ by $b$ this condition remains true. QED.
So, for example, the series $\sum\limits_{n=0}^{\infty}x^{-n^2}y^n$ doesn't lie in $K_0$. Here (https://math.stackexchange.com/a/2906674/204290) @Lubin proves that this series doesn't lie in $K_0$ by using Weierstrass preparation theorem.
