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Source. Theorem 3.2, preprint p. 11, its proof on p. 12; Proposition 3.2, p. 9, with its proof on pp. 9--11. Read on the rendered pages. The edition read is identified on the source card.

Statement

Let $(a_n)_{n\ge1}$ be a sequence of positive integers such that $a_N,a_{N+1},\ldots,a_{4N}$ form a geometric sequence for infinitely many $N$, and assume that

$$a_n=o\bigl(n^{n/7}\bigr)\qquad(14)$$

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

Proposition 3.2 (p. 9)

The theorem is derived from the proof of this proposition, with $a=1$, $b=0$ and positive terms. Let $a>0$ and $b\ge0$ be fixed integers and $(a_n)$ a sequence of Gaussian integers such that $a_N,\ldots,a_{4N}$ form a geometric sequence for infinitely many $N$, with $a_n=o(n^{n/7})$ for $n$ sufficiently large (9). Then either $S=\sum_{n\ge1}a_n/(a+b)_{a,n}\notin\mathbb{Q}[i]$, or

$$a^N\sum_{n=0}^{N-1}a_{2N+n}\frac{(N+n)!}{n!\,(2aN+b)_{a,N+n+1}}=o\bigl(N^{-N/8}\bigr)$$

for infinitely many such $N$. Here $(x)_{a,n}=x(x+a)\cdots(x+(n-1)a)$.

Proof pointer

Proposition 3.2 (pp. 9--11) assumes $S=t/q$ and forms, for large $N$ with a geometric run, a Gaussian integer $D_N$ from the tails with $N$-th difference weights. Lemma 2.3 reduces $D_N$ to the displayed main term plus $o(N^{-N/8})$, which bounds $|D_N|$ by $(N!)^{6/7}$ (12), while (13) makes $D_N$ divisible by $N!/A_N$ with $N!/A_N>(N!)^{6/7}$; so $D_N=0$. For Theorem 3.2 ($a=1$, $b=0$, positive terms) every term of the main sum is positive, and Stirling's formula gives $|D_N|>7^{-N}$ for large $N$ (p. 12), so $D_N\ne0$.

Dependencies

Proposition 3.2, Lemma 2.1 and Lemma 2.3 of the same paper. Theorem 4.2 applies this theorem: Theorem 4.2.

Bears on

No catalog problem directly.