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Source. Theorem 6.2 and its proof, preprint p. 12, with the sentence before it and the note after it on the same page; the introduction, p. 2. Read on the rendered pages.

Statement

Theorem 6.2 (p. 12): "Let $\{a_n\}_{n=1}^{\infty}$ be an unbounded monotonic sequence of positive integers. Then $S=\sum_{n=1}^{\infty}\frac{n}{a_1\ldots a_n}$ is rational if and only if $\frac{n}{a_n-1}$ is constant for $n\ge n_0$."

The paper adds after the proof (p. 12) that a bounded monotonic sequence of positive integers is constant from some $n_0$ on, so that the sum is then rational. Together the two statements decide the rationality of $\sum n/(a_1\cdots a_n)$ for every monotonic $a_n$; the introduction (p. 2) says that in Theorems 6.1 and 6.2 "the monotonicity of $\{a_n\}_{n=1}^{\infty}$ is crucial."

The rational case is explicit: if $n/(a_n-1)$ is a constant $c$ for $n\ge n_0$, then $a_n-1=n/c$ is an integer for two consecutive values of $n$, so $1/c$ is a positive integer $m$ and $a_n=mn+1$ from $n_0$ on.

Proof sketch (p. 12)

The paper introduces the theorem as proved by the method of Theorem 6.1, the primality of the numerators there being "used, but not crucial", and works out only the new features. Suppose $S=r/q$. Along $N$ with $a_{2N}<N^{0.7}$, an analog of the estimate (6) of Theorem 6.1 holds for numerators $n$ on most of $[N,2N)$, and there the tails can neither rise (that would need $a_N$ bounded in terms of $q$) nor fall (the numerators increase), while two equalities in a row force $a_{n+1}>a_n$ against the choice of $n$. When instead $a_n>n^{0.6}$ for all large $n$, (3) gives that a rise of the tails needs $a_n\le2q$, which happens for only finitely many $n$; the positive integers $qS_n$ can fall only finitely often, so the tails are eventually constant and $n/(a_n-1)$ equals that constant.

Bears on. No catalog problem directly; the analog of Theorem 6.1 with numerators $n$ in place of the primes.