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 $k$ ?


2 Answers 2


(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 $\{0,1\}$.

  • $\begingroup$ I just saw the answer of Lubin to this question link. He said that $\sum x^{2^n}$ is not rational over $Q$ by using some specific feature of $Q$, but can I generalize this to general field because the coefficients are only 0 and 1 and they satisfy the criterion on Hankel determinants? $\endgroup$
    – jlidm
    May 26 at 15:51

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.

  • $\begingroup$ Thanks a lot. I think I got what you mean. So in the same way, any power series $\sum x^{k_n}$, where $k_n$ tends to infinity is an example, right? $\endgroup$
    – jlidm
    May 26 at 16:04
  • $\begingroup$ @jlidm Well no, for example $k_n = n$ would not work. But if $k_{n+1} - k_n \rightarrow \infty$ (so we will have arbitrarily long gaps between nonzero coefficients) then it will work. $\endgroup$ May 26 at 16:08
  • $\begingroup$ Oh yes you're right, I made a mistake on that. $\endgroup$
    – jlidm
    May 26 at 16:09

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