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Source. Theorem 3.2 and its proof, preprint pp. 5--6; Proposition 3.1 and Lemma 3.1, pp. 4--5. Read on the rendered pages.

Statement

Theorem 3.2 (p. 5): "Let $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ be two sequences of integers such that $a_n\not|\,b_n$ for every $n$. Suppose

$$\liminf_{n\to\infty}\frac{|b_n|}{a_n}=0\qquad\text{and}\qquad \lim_{n\to\infty}\frac{b_n}{a_{n-1}a_n}=0 .$$

Then $S=\sum_{n=1}^{\infty}\frac{b_n}{a_1\ldots a_n}$ is irrational."

The standing convention (p. 3) applies: the $a_n$ are integers greater than $1$. The paper calls a series with $a_n\nmid b_n$ for every $n$ non-degenerate (p. 3); then every $b_n$ is nonzero. The introduction (p. 2) states the theorem in that language: a non-degenerate series is irrational if $\liminf|b_n|/a_n=0$ and $b_n=o(a_na_{n-1})$. The paper presents it as a variant of Corollary 3.2 (p. 5), Oppenheim's criterion, which assumes $|b_n|<a_n$ for every $n$ and $\liminf(|b_n|+1)/a_n=0$.

Proof sketch (pp. 5--6)

The proof rests on Proposition 3.1 (p. 4): for a non-degenerate series, $\liminf_{N\to\infty}|S_N|=0$ already forces $S$ to be irrational. That proposition combines Lemma 2.1 with Lemma 3.1, which says that a block $\sum_{n=M}^{N-1}b_n/(a_M\cdots a_n)$ cannot vanish when $a_{N-1}\nmid b_{N-1}$: a rational $S$ would make $qS_N$ integers tending to $0$ along a subsequence, hence two indices $M<N$ with $S_M=S_N=0$ and a vanishing block. Theorem 3.2 supplies the small tails: at an index $N$ where $|b_N|/a_N$ is small and every later $|b_n|/(a_{n-1}a_n)$ is small, the tail $S_N$ is bounded by its first term plus a geometric series, so it is small.

Uses

The proof of Theorem 5.1 (p. 9) appeals to Theorem 3.2 in the case where the tails decrease at more than half the indices of a dyadic range, after showing that $a_n$ then grows fast.

Bears on. No catalog problem directly; a tool in the proof of Theorem 5.1.