16
$\begingroup$

For $A \subseteq \mathbb{N}$, define $\displaystyle x_A = \sum_{n \in A} \frac{1}{n!}$. It is easy to see that for every infinite $A$, $x_A$ is irrational.

Question: Is there an infinite $A \subseteq \mathbb{N}$ for which $\displaystyle x_A = \sum_{n \in A} \frac{1}{n!}$ is algebraic?

Improve this question
$\endgroup$
4
  • 5
    $\begingroup$ Exactly this class of numbers is proved irrational (and some examples given, which I think all happen to be transcendental) in Martin Griffiths, "Irrational sums from reciprocals of factorials", Math. Gaz. 2015, jstor.org/stable/24496963 $\endgroup$
    – David Eppstein
    Commented Apr 3, 2022 at 7:35
  • 3
    $\begingroup$ I suspect the answer is no but that a proof is beyond current technology. The proof that $e$ is transcendental relies on special properties of the function $e^x$, not just on the rate of growth of $n!$. There are transcendence results that rely only on how fast the summands shrink but they require a faster growth rate than $n!$. For example, Nyblom proves that if for some fixed $\lambda>2$ we have $\liminf_{n\to\infty} a_{n+1}/a_n^{λ+1}>1$ then $\sum_n 1/a_n$ is transcendental. $\endgroup$
    – Timothy Chow
    Commented Apr 4, 2022 at 2:33
  • $\begingroup$ For other results along these lines, see Kumar and Vance and the references therein. I'm not familiar with them all but at first glance they all require faster growth rates than $n!$. And for an arbitrary $A$ it seems to me that there's not much you can say about $\sum_{n\in A} 1/n!$ other than the general rate at which the summands shrink. $\endgroup$
    – Timothy Chow
    Commented Apr 4, 2022 at 2:34
  • $\begingroup$ At least we can prove that if $x_{A}$ is an algebraic number then $A$ is not a cofinite subset of $\mathbb{N}$. :) $\endgroup$
    – José Hdz. Stgo.
    Commented Apr 4, 2022 at 18:15

0

Reset to default

You must log in to answer this question.