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?
Best Answer
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.