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Source. Theorem 3.1 and its proof, preprint pp. 7--8; Corollary 3.1, p. 8; Proposition 3.1, p. 5, with its proof on pp. 5--6. Read on the rendered pages. The edition read is identified on the source card.

Statement

Let $0<\delta<1$, let $(R(n))_{n\ge1}$ be a sequence of positive integers and $(a_n)_{n\ge1}$ a sequence of integers. Suppose that for infinitely many $N\in\mathbb{N}$ the terms

$$a_{N-2R(N)},\ a_{N-2R(N)+1},\ \ldots,\ a_{\lceil N+5R(N)/(1-\delta)\rceil}$$

form a geometric progression, and that

$$a_{N+n}=o\bigl(N^{R(N)+\delta n}\bigr)\quad(n=0,1,\ldots) \qquad\text{and}\qquad N-2R(N)\to\infty\quad(N\to\infty).$$

Then $S=\sum_{n=1}^{\infty}a_n/n!\notin\mathbb{Q}$.

Here $\lceil x\rceil$ is the least integer at least $x$ (p. 7).

Corollary 3.1 (p. 8) is the case $\delta=1/6$: with $(R(n))$ and $(a_n)$ as above, if for infinitely many $N$ the terms $a_{N-2R},\ldots,a_{N+6R}$ form a geometric progression and $a_{N+n}=o(N^{R(N)+n/6})$ for $n=0,1,\ldots$, with $N-2R(N)\to\infty$, then $S\notin\mathbb{Q}$.

Scope of the hypothesis (an observation of this page, not of the paper). The proof writes the ratio as $a_{N+1}/a_N=c/d$ with $c$, $d$ coprime positive integers, so it covers only runs of nonzero terms with positive ratio. A run of zeros meets the printed wording, and a sequence that is zero from some index on gives a rational $S$, so zero runs must be excluded. Remark 3.1 (p. 7) says the argument also works for a negative ratio $c/d$ under the extra inequality $-\frac cd\cdot\frac{K+1}{a(N+K+1)+b}<\frac12$, in the notation of Proposition 3.1 below.

Proof pointer

The theorem is derived from Proposition 3.1 (p. 5). That proposition takes integers $a>0$, $b\ge0$ and integers $a_n$ such that for infinitely many $N$ the run $a_{N-K},\ldots,a_{N+K+H}$ is a nonzero geometric progression with ratio $c/d$, $c=c(N)$ and $d=d(N)$ coprime positive integers, $K=K(N)$, $H=H(N)$, $N-K\to\infty$; under the two growth conditions (5) and (6) it concludes that $\sum_n a_n/(a+b)_{a,n}\notin\mathbb{Q}$, where $(x)_{a,n}=x(x+a)\cdots(x+(n-1)a)$. The proof forms an integer combination $D_N$ of tails of the series with $K$-th difference weights, evaluates it with the summation identity of Lemma 2.3 (p. 4), and plays a lower bound for $|D_N|$ against its divisibility by $K!/A_K$. Theorem 3.1 takes $a=1$, $b=0$, $K=2R$ and $H=\lceil\frac{3+2\delta}{1-\delta}R\rceil$ with $R=R(N)$, and checks (5) and (6) from the growth bound (pp. 7--8).

Dependencies

Proposition 3.1, Lemma 2.1 and Lemma 2.3 of the same paper.

Bears on

No catalog problem directly. It is the tool behind Corollary 3.2.