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Statement

Theorem 1 (p. 1). Let $n_1<n_2<\cdots$ be an infinite sequence of integers such that

$$\limsup_{k\to\infty}n_k^{1/2^k}=\infty \qquad(2)$$

and

$$n_k>k^{1+\epsilon} \qquad(3)$$

for some fixed $\epsilon>0$ and every $k>k_0(\epsilon)$. Then

$$\alpha=\sum_{k=1}^\infty\frac1{n_k}$$

is irrational. The equation numbers (2) and (3) are the paper's, and later statements of the paper refer to them.

Sharpness (p. 2, without proof). The paper calls Theorem 1 best possible in both hypotheses.

• For (2): it calls it well known and easy that for every $A$ there is a sequence with $n_k>A^{2^k}$ for every $k>0$ and $\sum_k1/n_k$ rational.
• For (3): if $f(k)\to\infty$ and $\log f(k)/\log k\to0$, there is a sequence satisfying (2) and $n_k>kf(k)$ for all $k$ with $\sum_k1/n_k$ rational. The paper leaves the details to the reader.

Variant (p. 6, unnumbered). After the proof of Theorem 1 the paper states that the same method easily proves that $\sum_k1/n_k$ is irrational if $\liminf_{k\to\infty}n_k^{1/2^k}>1$ and $\lim_{k\to\infty}n_k^{1/2^k}$ does not exist. The sentence does not restate the other hypotheses of Theorem 1, and no proof is printed.

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

Read depth. Claims checked: the statement, the sharpness remarks and the variant were read clause by clause on the printed pages. The proof (pp. 3--6) was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 3--6. The unnumbered Lemma (p. 3) says that if $n_1<n_2<\cdots$ satisfy (3) for every $k$, then the tail $\sum_{i\ge1}1/n_{k+i}$ is less than $c_\epsilon/n_{k+1}^{\epsilon/(1+\epsilon)}$; it follows from counting the $n_i$ below $x$ through (3). With $M_k=n_1\cdots n_k$ and $\alpha=a/b$, the number $bM_k\sum_{i\ge1}1/n_{k+i}$ is a positive integer, so it is at least $1$. The proof splits into three cases.

1. If for every $l$ some $k$ has $n_{k+1}>M_k^l$ (the paper's (9)), the Lemma makes that integer less than $1$ once $l>(1+\epsilon)/\epsilon$ and $k$ is large.
2. Otherwise some $l$ has $n_{k+1}<M_k^l$ for every $k$, which gives $n_k<2^{(l+1)^k}$. If moreover $n_k>2^k$ for every $k>k_0$, the tail is at most a constant times $\log n_k/n_k$, and (2) supplies infinitely many $k$ at which $L_k=n_k^{1/2^k}$ exceeds $(1+1/k^2)$ times all earlier values, an idea the paper credits to Borel; at such $k$ the integer bound forces $n_{k+1}$ to grow faster than the bound $2^{(l+1)^k}$ allows.
3. If $n_k\le2^k$ for infinitely many $k$, the paper shows that $\liminf_kM_k\sum_{i\ge1}1/n_{k+i}=0$, choosing the indices from (2), the Lemma and the bounds of the previous case.

Dependencies

The unnumbered Lemma of the same paper (p. 3). The paper's first page also quotes an earlier theorem of Erdős and Straus (its reference [1]), which the proof does not use.

Bears on

No Erdős problem is stated in terms of this theorem. The paper's Theorem 3 uses it to reduce to the case $\limsup m_k^{1/2^k}<\infty$; see Theorem 3 for Problem 262.