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Source. Theorem 4.1, preprint p. 13, its proof on pp. 13--14; Corollary 4.1 and the remark before it, p. 14. Read on the rendered pages. The edition read is identified on the source card.

Statement

Let $m>1$ be a fixed integer and $(b_n)_{n\ge1}$ a sequence of positive integers such that

$$\liminf_{n\to\infty}\frac{b_n}{n}<\frac1{m-1}\qquad(16)$$

and

$$m^{b_n-b_{n-1}}<\frac n2\qquad(17)$$

for all large $n$. Then $\sum_{n=1}^{\infty}m^{b_n}/n!\notin\mathbb{Q}$.

Corollary 4.1 (p. 14). Let $m$ be a positive integer and $A$ an infinite subset of $\mathbb{N}$ with lower asymptotic density less than $1/(m-1)$, and let $b_n$ be the number of elements of $A$ that are at most $n$. Then $S=\sum_{n=1}^{\infty}m^{b_n}/n!\notin\mathbb{Q}$. The printed proof is that (16) and (17) hold.

Reading notes (observations of this page, not of the paper). The density bound $1/(m-1)$ and Theorem 4.1 both need $m\ge2$; for $m=1$ the sum is $e-1$, irrational for the classical reason. The remark before the corollary says it gives the irrationality of $\sum m^{\pi(n)}/n!$ for $m=0,1,2,\ldots$ (p. 14); for $m\ge2$ this is the corollary with $A$ the primes, of density $0$, while for $m=0$, with $0^0=1$, the sum is $1$, so that case of the remark does not stand.

Proof pointer

Pages 13--14. Assuming the sum is $t/q$, the integer $tN!/q-\sum_{n\le N}N!\,m^{b_n}/n!$ equals a positive tail that (17) keeps below $m^{b_N}$, while it is divisible by a power of $m$ whose exponent is governed by $[N/m]+[N/m^2]+\cdots$; condition (16) makes $-b_n+[n/m]+[n/m^2]+\cdots$ unbounded above, which gives arbitrarily large $N$ contradicting the inequality (18).

Dependencies

Lemma 2.1 of the same paper.

Bears on

No catalog problem directly.