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Statement

Throughout Section 2 the series is

$$\xi=\sum_{n=1}^{\infty}\frac{d(n)}{a_1a_2\cdots a_n},$$

the paper's (2.1), with $d(n)$ the number of divisors of $n$ and the $a_n$ positive integers with $2\le a_1\le a_2\le\cdots$ (p. 638).

Lemma 2.2 (p. 638). "The series (2.1) is irrational if there exists a $\delta>0$ so that the inequality $a_n<(\log n)^{1-\delta}$ holds for infinitely many values of $n$."

The section's monotonicity convention is a hypothesis here: the proof uses it at its first step (p. 638).

Source. P. Erdős, E. G. Straus, Some number theoretic results, Pacific J. Math. 36 (1971), no. 3, 635--646; Lemma 2.2 on p. 638, its proof on pp. 638--640. The copy read is identified on the source card.

Read depth. Claims checked: the statement and the convention (2.1) were read on the page image of p. 638; the proof was read for structure, not checked. Nothing here is independently reviewed.

Proof pointer

Pages 638--640; the paper says this case is very similar to Erdős's 1948 proof for $\sum d(n)t^{-n}$. For a large $n$ with $a_n$ small, monotonicity gives an interval $I$ of length $n/\log n$ below $n$ on which $a_i$ is a constant $t$. With $k=[(\log n)^{\delta/10}]$ and primes above $(\log n)^2$, a Chinese-remainder system (2.4) makes $d(x+i)$ divisible by $t^{i+1}$ for $0\le i<k$, so if $\xi=a/b$ the first $k$ terms after $x$ contribute an integer. Divisor-sum averaging over the solutions in $I$ (2.12) finds one where the next $10\log n$ divisor values are below $2^{k/4}$, and the remaining tail is then less than $1$ (2.7), (2.13), a contradiction.

Dependencies

The Chinese remainder theorem; the elementary bound (2.12) on sums of $d$ over an arithmetic progression. The method is that of P. Erdős, On arithmetical properties of Lambert series, J. Indian Math. Soc. 12 (1948), 63--66 (the paper's reference [1]).

Bears on

• #258: one of the two cases joined in Theorem 2.23, the nondecreasing case; it concerns only nondecreasing sequences.