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Statement

Property P (pp. 2--3). Following a question of Erdős and Straus, prompted by the identity $\sum_{n\ge0}1/((n+2)\,n!)=1$, a sequence $n_1<n_2<\cdots$ has property $P$ if $\sum_k1/m_k$ is irrational for every choice of integers $m_k>0$ with $m_k\equiv0\pmod{n_k}$. The paper reports that they wondered whether $n_k=2^{2^k}$ has property $P$, and announces a proof.

Theorem 3 (p. 6). If $n_k>0$ and $n_k\equiv0\pmod{2^{2^k}}$ for every $k\ge1$, then $\alpha=\sum_{k=1}^\infty1/n_k$ is irrational. The paper stresses that the sequence $n_k$ is not assumed monotone. Hence $2^{2^k}$ has property $P$.

Remarks on property P (p. 3 and p. 7).

• The paper states that property $P$ is only interesting if $\lim n_k^{1/2^k}<\infty$.
• It states that it cannot prove that a sequence with property $P$ of that kind exists if pairwise coprimality, $(n_i,n_j)=1$, is also assumed.
• It does not know whether some sequence with property $P$ has $n_k$ not tending to infinity very fast.
• The paper closes (p. 7, quoted): "I cannot decide whether there is a sequence $u_k$ having property $P$ and satisfying $u_k^{1/2^k}\to1$, or $u_k>C^{2^k}$, $(u_i,u_j)=1$. I would tentatively guess that such sequences exist."

Source. P. Erdős, Some problems and results on the irrationality of the sum of infinite series, J. Math. Sci. 10 (1975), 1--7: property $P$ on pp. 2--3, Theorem 3 and its proof on pp. 6--7, the closing remark on p. 7. The edition read is identified on the source card.

Read depth. Claims checked: the definition, the statement and the remarks were read clause by clause on the printed pages. The proof (pp. 6--7) was read but not checked step by step. The sentence after display (30) names $N_k$ the least common multiple of $m_1,\ldots,m_k$ and $M_k$ their product, but display (30), and the use of the two letters in (32), (33) and the final inequality on p. 7, fit only the reverse reading ($M_k$ the least common multiple, $N_k$ the product); the pointer below avoids both letters. Nothing here is independently reviewed.

Proof pointer

Pages 6--7. Reorder the $n_k$ as a nondecreasing sequence $m_1\le m_2\le\cdots$; then $m_k\ge2^{2^k}$. If $\limsup m_k^{1/2^k}=\infty$, Theorem 1 gives irrationality, so one may assume the limsup is a finite $C$. As in the paper's Lemma, the tail after $m_k$ is then bounded by a constant over a single term (the paper's (29)). Among $m_1,\ldots,m_k$ at least two are divisible by $2^{2^{k-1}}$, so the least common multiple of $m_1,\ldots,m_k$ is at most their product divided by $2^{2^{k-1}}$. Along indices $k_r$ with $m_{k_r}>(C-\epsilon_r)^{2^{k_r}}$ and $\epsilon_r\to0$ (the paper's (31)), this saving makes the least common multiple of $m_1,\ldots,m_{k_r-1}$ times the tail from $m_{k_r}$ on tend to $0$ (the paper's (32)), which a rational $\alpha$ forbids.

Dependencies

Theorem 1 and the unnumbered Lemma (p. 3) of the same paper.

Bears on

• Problem 262: property $P$ is the problem's notion, with $m_k=t_ka_k$ and $t_k\ge1$, and Theorem 3 shows that $a_n=2^{2^n}$ is such a sequence, so a sequence of this kind can grow as slowly as $2^{2^n}$. The paper does not show that slower sequences fail; it states that it does not know. The claim page for this theorem records it on the problem.