Explicit examples of formal power series which is not rational functions? This MO question https://mathoverflow.net/questions/249541/formal-power-series-is-taylor-expansion-of-rational-function-iff-hankel-determin states that if $k$ is a field and $k[[T]]$ is the power series ring over $k$ then $u(T)\in k[[T]]$ is the power series of some rational function if and only if a condition on Hankel determinant holds. However, I could not think of any explicit power series which does not satisfy this condition.
I know that if $k=\mathbb{R},\mathbb{C}$, there are a lot of examples of power series which are not rational functions. But are there any explicit examples of power series which work for an arbitrary field $k$ ?
 A: (1) A lacunary series like $\sum_{j=0}^{\infty} T^{j!}$ is not a rational function.  You will get many Hankel determinants with $1$ on the anti-diagonal (upper right to lower left) and $0$ everywhere else.
(2)  If $k$ is a countable field, than there are countably many rational functions, but uncountably many power series $\sum_{j=0}^\infty a_j T^j$ with coefficients in $\{0,1\}$.
A: It's easiest to use condition (2):
There is a finite sequence $q_0,\ldots, q_N$, not all zero, such that for all sufficiently large $m$,
$$a_m q_N + a_{m+1} q_{N-1} + \cdots + a_{m + N}q_0 = 0.$$
In other words, the coefficients must obey a linear recurrence (with a finite number of exceptions.)
If you choose coefficients "sufficiently random" you would not expect this to hold.  For an explicit counterexample, you can take $f(T) = \sum_{n=0}^\infty a_n T^n$ where $a_n = 1$ if $n = 2^m$ for some $m$, otherwise $a_n = 0$.  Then you will have arbitrarily long "gaps" in the sequence, with only coefficients equal to $0$.  If the recurrence relation is obeyed, then any $N$ consecutive zeros $a_m, \ldots, a_{m+N-1}$ would mean that all coefficients after must be $0$, since you can solve for $a_{m+N}$ as a linear combination of $a_{m}, a_{m+1}, \ldots, a_{m+N-1}$, so that $a_{m+N} = 0$, and so on infinitum.  Then we conclude $f(T)$ must be a polynomial, which we know it is not. So no such recurrence can hold.
A: A good class of examples is algebraic power series that are not rational. For example, consider the Catalan number generating function
$$c(x) = \frac{1-\sqrt{1-4x}}{2x}= \sum_{n=0}^\infty \frac{1}{n+1}\binom{2n}{n}x^n$$
which is defined over every field. It satisfies $c(x) = 1+xc(x)^2$, and it's not hard to show from this that $c(x)$ is not rational.
