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Source. Theorem 2.1, printed pp. 85--86; proof pp. 86--87; Remark on p. 87. Read on the page images.

Statement

Take integers $b_n$ and positive integers $a_n$ such that $a_n\ge2$ once $n$ is large and

$$\lim_{n\to\infty}\frac{|b_n|}{a_{n-1}a_n}=0\qquad(2.2)$$

(printed with "$n=1$" under the limit, a misprint for $n\to\infty$). The sum

$$\sum_{n=1}^{\infty}\frac{b_n}{a_1\cdots a_n}\qquad(2.3)$$

is a rational number exactly when some positive integer $B$ and some integers $c_n$ satisfy, for every sufficiently large $n$,

$$Bb_n=c_na_n-c_{n+1},\qquad|c_{n+1}|<a_n/2 .\qquad(2.4)$$

Proof structure (pp. 86--87)

Sufficiency. If (2.4) holds beyond $N$, then $Ba_1\cdots a_{N-1}\sum_{n\ge1}b_n/(a_1\cdots a_n)$ equals an integer plus $\sum_{n\ge N}(c_na_n-c_{n+1})/(a_N\cdots a_n)$, which telescopes to $c_N$; so the series is rational.

Necessity. Let the sum (2.3) be $A/B$, and take $N$ with $a_n\ge2$ and $|b_n/(a_{n-1}a_n)|<1/(4B)$ whenever $n\ge N$. Then (2.5) $Aa_1\cdots a_{N-1}=\text{integer}+Bb_N/a_N+R_N$ with $R_N=\sum_{n>N}Bb_n/(a_N\cdots a_n)$ and (2.6) $|R_N|<1/2$. With $c_N$ the integer nearest to $Bb_N/a_N$ and $c_{N+1}$ defined by $Bb_N=c_Na_N-c_{N+1}$, (2.5) makes $-c_{N+1}/a_N+R_N$ an integer smaller than $1$ in absolute value, so it vanishes; this gives (2.7)--(2.8) $Bb_{N+1}/a_{N+1}=c_{N+1}-R_{N+1}$, so $c_{N+1}$ is the integer nearest to $Bb_{N+1}/a_{N+1}$, and the construction continues.

Remark (p. 87). As (2.2) makes the tails $R_n$ tend to $0$, a rational sum forces $c_{n+1}/a_n\to0$; so either $a_n\to\infty$, or from some point on $c_n=0$ and therefore $b_n=0$.

Role

The criterion behind Corollary 2.10, Theorem 3.1 and Theorem 3.7. For $a_n=n$ it is a criterion for factorial series with $b_n=o(n^2)$; the later exact tests of Tijdeman–Yuan 2002 and Hančl–Tijdeman 2004 work from increments $b_{n+1}-b_n$ instead. The paper says the section modifies [2, Lemma 2.29] of the authors' 1971 paper. The Archive of Formal Proofs entry named on the card reports a formalization of this theorem; not inspected here.

Bears on. No catalog problem directly; it is the tool behind the results linked above.