---

Statement

The paragraph opening p. 131 asks how far conditions (i) and (ii) of Theorem 1 are necessary, and says that finiteness of $\limsup n_k^2/n_{k+1}$ does not suffice in place of (i). Two examples follow; $a$ is a positive integer, and the examples are relevant when $a\ge2$.

First example. Take $\sum1/(an_k)$, where $\sum1/n_k$ is Sylvester's series (1). The sum is $1/a$, the denominators $an_k$ satisfy $\lim(an_k)^2/(an_{k+1})=a$, and condition (ii) still holds.

Second example. The paper takes the series $1/a=\sum1/n_k$ with $n_1=a+1$, its odd-indexed terms chosen greedily (each $n_{2k+1}$ least with $1/n_1+\cdots+1/n_{2k+1}<1/a$) and its even-indexed terms by a modified greedy rule (each $n_{2k}$ least with $1/n_1+\cdots+1/n_{2k-1}+1/(n_{2k}-a+1)<1/a$). It states that then $n_{2k+1}\equiv1\pmod a$, $n_{2k}\equiv0\pmod a$,

$$n_{2k}=n_{2k-1}^2-n_{2k-1}+a,\qquad n_{2k+1}=n_{2k}^2/a-n_{2k}+1,$$

so that $\lim n_{2k-1}^2/n_{2k}=1$ and $\lim n_{2k}^2/n_{2k+1}=a$, giving $\limsup n_k^2/n_{k+1}=a$ and $\liminf n_k^2/n_{k+1}=1$, and that $N_k/n_{k+1}\le a^{-[k/2]+1}n_1\cdots n_k/n_{k+1}$ is bounded. The paper adds that the construction could easily be modified so that $\{n_k\}$ satisfies no algebraic recursion relation.

These are the paper's claims, stated with only the computation shown; the congruences and recurrences were not rederived here.

Dependencies

Sylvester's series (1) of Theorem 1.

Source. P. Erdős and E. G. Straus, On the irrationality of certain Ahmes series, J. Indian Math. Soc. (N.S.) 27 (1964), 129--133; the edition read is named on the source card.

Read depth. Claims checked: the paragraph was read clause by clause on the page image of p. 131. Nothing here is independently reviewed.

Bears on

• Problem 243: background only. Both examples have $\limsup n_k^2/n_{k+1}=a$, so for $a\ge2$ they fail the problem's hypothesis $a_n/a_{n-1}^2\to1$. They show that condition (i) of Theorem 1 cannot be replaced by finiteness of $\limsup n_k^2/n_{k+1}$.