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Statement

Theorem 2 (pp. 1--2). Let $n_1<n_2<\cdots$ be an infinite sequence of integers, as in Theorem 1, satisfying (3), that is $n_k>k^{1+\epsilon}$ for some fixed $\epsilon>0$ and every $k>k_0(\epsilon)$, and suppose that for every $t$

$$\limsup_{k\to\infty}n_k^{1/t^k}=\infty. \qquad(4)$$

Then $\alpha=\sum_{k=1}^\infty1/n_k$ is a Liouville number.

Sharpness (p. 2). The paper observes that $\sum_k1/2^{2^k}$ is not a Liouville number, so (4) is best possible. It adds that it expects a much weaker condition than (3) to suffice together with (4), and that it has not settled this.

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

Read depth. Claims checked: the statement and the sharpness remark were read clause by clause on the printed pages. The proof (p. 3) was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Page 3. With $M_k=n_1\cdots n_k$ it suffices to find, for every $s$, a $k$ with $\bigl|\alpha-\sum_{i\le k}1/n_i\bigr|<1/M_k^s$ (the paper's (6)). For a large $t$ depending on $\epsilon$ and $s$, condition (4) gives a $k$ at which $n_{k+1}^{1/t^{k+1}}$ exceeds $n_j^{1/t^j}$ for every $j\le k$; then $M_k<n_{k+1}^{1/(t-1)}$, and the paper's unnumbered Lemma (p. 3), which bounds the tail $\sum_{i\ge1}1/n_{k+i}$ by $c_\epsilon/n_{k+1}^{\epsilon/(1+\epsilon)}$ under (3), gives (6).

Dependencies

The unnumbered Lemma of the same paper (p. 3), stated on the Theorem 1 page.

Bears on

• Problem 247: an observation of this page, not of the paper. Taking $n_k=2^{a_k}$ for an increasing sequence of positive integers $a_k$, condition (3) holds because $2^{a_k}\ge2^k$, and (4) says that $\limsup_ka_k/t^k=\infty$ for every $t$. For such sequences Theorem 2 makes $\sum_k2^{-a_k}$ a Liouville number, hence transcendental. This covers only sequences that grow faster than every exponential along a subsequence, a small part of the problem's hypothesis $\limsup a_n/n=\infty$; it decides no other instance.