TITLE: Explicit elements of $K((x))((y)) \setminus K((x,y))$
QUESTION [40 upvotes]: In an answer to the popular question on common false beliefs in mathematics
Examples of common false beliefs in mathematics
I mentioned that many people conflate the two different kinds of formal Laurent series field in two variables.  Let $K$ be any field.  Then we have the joint Laurent series field, the field of fractions
$K((x,y)) = \operatorname{Frac}(K[[x,y]])$ 
and also the iterated Laurent series field 
$K((x))((y)) = \left(K((x))\right)((y))$.  
These fields are not the same, roughly because when we write out an arbitrary element of the iterated field as a formal Laurent series in $y$, for each (say non-negative) $n$, the coefficient of $y^n$ is allowed to be an arbitrary formal Laurent series in $x$.  In particular, as $n$ varies, arbitrarily large negative powers of $x$ may appear.  
However, this is rather far from a convincing argument.  Indeed, I gave the following explicit (and fallacious!) example: $\sum_{n=0}^{\infty} x^{-n} y^n$.  But in a comment to my answer, user AS points out that this element is equal (in the iterated field, say) to $\frac{1}{1-\frac{y}{x}}$ and therefore it must lie in the fraction field of $K[[x,y]]$.  Evidently the fallacy here is that the fraction field of $K[[x,y]]$ is the field of all formal Laurent series which are finite-tailed in both $x$ and $y$.  But as this example shows, the latter isn't even a field, unlike the one-variable case.
[At least when $K = \mathbb{C}$, by less explicit means one can see that these two fields are very different: e.g., the joint field is Hilbertian so has nonabelian Galois group, whereas the iterated field has Galois group $\widehat{\mathbb{Z}}^2$.]
AS offered to write down, with proof, an explicit element of $K((x))((y)) \setminus K((x,y))$, so I decided to post a question asking for such a guy.  Of course, there is more than one such element -- or better put, more than one type of construction of such elements -- so I would be interested to see multiple answers to:

Please exhibit, with proof, an explicit element of $K((x))((y)) \setminus K((x,y))$.

REPLY [6 votes]: Addendum: actually there is a chain of subfields between $K((x,y))$ and $K((x))((y))$ with cardinality c.
For instance, for any $\lambda>1$ we may consider the subset $R_\lambda$ of $K((x))((y))$ of all Laurent series $\sum_k c_k(x)y^k$ with $c_k\in K((x))$ satisfying $$\inf_k \lambda ^{-k} \mathrm{ord}(c_k)   > -\infty.$$
It's easy to check that it's a subfield of $K((x))((y))$, containing $K[[x,y]]$.  
Moreover, since in place of $\lambda^k$ we can use functions $\mathbb{N}\to\mathbb{N}$ with arbitrarily large growth, one can also show that the subfields between $K((x,y))$ and $K((x))((y))$ have uncountable cofinality also.