David Roberts wrote in the comment section of the blog post "Convergence of an infinite sum in the rationals" the following paragraph:

Someone mentioned (I think on Twitter) that the Taylor series of rational functions should all be like this example (which is easy to see), but possibly also that this is the only class of power series that converges like this in the rationals, namely, if a power series converges on the rationals, then it is the Taylor series for a rational function. Not sure how one would show this.

Note that David Roberts is working inside of the rational numbers $\mathbb{Q}$, rather than the real numbers $\mathbb{R}$, in his blog post.

Is it true that in the rational numbers every convergent power series on the rational numbers is a Taylor series for a rational function on the rational numbers? If so, how would one go about proving this statement? If not, what counterexamples exist out there?

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    $\begingroup$ I'm not sure if this is what you're asking, but there are power series $f(x)\in\mathbb Q[[x]]$ that (1) converge for all $x\in\mathbb R$ and (2) for every $x\in\mathbb Q$, the value $f(x)$ is in $\mathbb Q$ and (3) $f(x)$ is not equal to a function in $\mathbb Q(x)$, and indeed, I think there are examples where $f(x)$ is transcendental over $\mathbb Q(x)$. My recollection is that there are examples of such series due to Mahler, but I don't recall the exact reference. $\endgroup$ Sep 7, 2022 at 17:00

1 Answer 1


No. Enumerate the rational numbers $a_1,a_2,\dots$. Then for every sequence $c_1, c_2,\dots$ of rational numbers decreasing rapidly enough, the series

$$ \sum_{n=1}^{\infty} c_n x^n \prod_{i=1}^{n-1} (x-a_i ) $$

converges on each rational number. On $a_m$ it takes the value $$ \sum_{n=1}^{m} c_n a_m^n \prod_{i=1}^{n-1} (a_m-a_i ) $$ which is rational.

By "decreasing rapidly enough", it suffices to have $$ \sum_{n=1}^{\infty} c_n |x|^n \prod_{i=1}^{n-1} ( |x| + |a_i|)< \infty $$ for each rational $x$, e.g. it suffices to have

$$|c_n| < \frac{1}{ n^n \prod_{i=1}^{n-1} ( n + |a_i|)}$$ as then for $|x|<m$, for all $m \geq n$, the $n$'th term in the above sequence is bounded by $(x/m)^n$ and thus that sequence converges.

There are uncountably many series of this type, so they can't all come from rational functions.

  • $\begingroup$ This is very nice! $\endgroup$ Sep 7, 2022 at 17:12
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    $\begingroup$ Indeed a very nice explanation. But as I noted in my comment, such examples have been around for quite a while. A very interesting question (to me) is how fast the coefficients have to decrease for such functions. For example, is it possible to create such a function with the $n$th coefficients decreasing only exponentially with $n$, instead of decreasing like $1/n^n$? $\endgroup$ Sep 7, 2022 at 18:52
  • $\begingroup$ @JoeSilverman Such a power series would not converge on all the rational numbers, right, rather on just an interval in them? It seems to me that one can just apply a similar construction, restricted to rational numbers on an interval. $\endgroup$
    – Will Sawin
    Sep 9, 2022 at 15:08
  • $\begingroup$ @JoeSilverman Another question is whether one can make a power series like this that doesn't converge everywhere, but whose analytic continuation still takes rational values on every rational. But then one can just take a Mahler example and add $1/(1-x)$. $\endgroup$
    – Will Sawin
    Sep 9, 2022 at 15:10
  • $\begingroup$ @WillSawin Good point regarding exponetial decay of coefficients. But there's room between exponential decay and $n^n$ decay. Suppose there is such a series (everywhere convergent, rational values at rational arguments) with $|c_n|\ge2^{-nf(n)}$. There do exist such series with $f(n)\asymp\log n$. How about with $f(n)\asymp(\log n)^{1-\epsilon}$? Or $f(n)\asymp(\log n)/(\log\log n)$? $\endgroup$ Sep 9, 2022 at 18:32

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