The irrationality of rapidly converging series

Question

I recently saw a pretty elegant proof of the irrationality of $e$, namely:

Let $s_n:=\sum_{k=0}^{n}{\frac{1}{k!}}$ such that $e=\lim_{n\to\infty} s_n$. We obviously have $s_n<e$ and furthermore $e=s_n+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...=s_n+\frac{1}{n!}\left(\frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}+...\right)<s_n+\frac{1}{n!}\left(\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+...\right)=s_n+\frac{1}{n!n}$. Therefore, we have $n!s_n<n!e<n!s_n+\frac{1}{n}$. If $e$ was rational, then $e=\frac p q$ for $p,q\in\mathbb N$. But since we can choose $n$ arbitrarily high we can assume $n>q$. Then the numbers $n!s_n$ and $n!e$ are clearly integers and we get $n!s_n<n!e<n!s_n+\frac{1}{n}≤n!s_n+1$, contradiction.

I adapted this proof to show the irrationality of numbers like $\sum_{k=0}^{\infty} \frac{1}{2^{n^2}}$ which for me raised the following question: This kind of proof seems to be working well with rapidly converging series. Can there be said anything rigorous about the irrationality of such series in order to justify this observation?

Answer 1

I would say that $\sum_0^\infty \frac{1}{n!}$ is irrational for a "different" reason than $\sum_0^\infty \frac{1}{2^{n^2}}$, and rapidity of convergence is not the full reason for the irrationality of either sum.

Let $a_0,a_1,a_2,a_3,\dots$ be a sequence of rationals that is very rapidly converging to $0$. Let $b_i=a_{i}-a_{i+1}$. Then the sequence $b_0,b_1,b_2, b_3,\dots$ converges very rapidly to $0$, and yet $\sum_{i=0}^\infty b_i$ is, by telescoping, equal to the rational $a_0$.

Answer 2

Indeed there can. According to this paper (pdf download can be found here) a sufficient condition for irrationality is

$$a_n-a_{n-1}^2+a_{n-1}-1>0$$

for all but a finite number of $n$. That is, if this condition holds (except for a finite number of cases) then

$$\sum_{n=1}^\infty \frac{1}{a_n}$$

is irrational. The paper describes that this is the "best possible" result since defining $a_n$ by the recurrence relation (with $a_1\in\{2,3,...\}$)

$$a_n=a_{n-1}^2-a_{n-1}+1$$

will produce a sum with a rational value.