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Source. Theorem 3.1 and its proof, preprint p. 4. Read on the rendered page.

Statement

Theorem 3.1 (p. 4): "If $b_n>0$, $b_{n+1}<(1+\epsilon)b_n$ for some $\epsilon<1$ and all $n\ge n_1$, and $\liminf_{n\to\infty}\frac{b_n}{a_n}=0$, then $S$ is irrational." Here $S=\sum_{n\ge1}b_n/(a_1\ldots a_n)$ under the standing convention: $a_n$, $b_n$ integers with $a_n>1$ for all $n$.

Proof sketch (p. 4)

The proof is a geometric-series bound on the tail $S_N$ of (2). If $S=r/q$, choose $N\ge n_1$ with $b_N/a_N$ below $(1-\epsilon)/(2q)$. From $N$ on each term of $S_N$ is less than $(1+\epsilon)/2$ times the term before it, because the numerators grow by a factor below $1+\epsilon$ and every $a_n$ is at least $2$. So $S_N$ is less than $b_N/a_N$ times $2/(1-\epsilon)$, that is, less than $1/q$. Lemma 2.1 makes $qS_N$ an integer, so $S_N=0$, which is impossible since every $b_n$ is positive.

Uses

Example 3.1 applies it with $b_n=p_n^k$ and $a_n=2^{p_n-p_{n-1}}$. Corollary 3.1 (p. 5) is the signed version ($a_n\nmid b_n$, $|b_{n+1}|<(1+\epsilon)|b_n|$ for some $\epsilon$ with $0<\epsilon<1$ and all $n\ge n_0$, $\liminf|b_n|/a_n=0$).

Bears on. No catalog problem directly; the tool behind Example 3.1.