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Statement

Example (pp. 205--206). Let $S=(4,1,3,3^2,3^3,\ldots,3^n,\ldots)$, so that $M(S)=(1,3,4,3^2,4\cdot3,3^3,4\cdot3^2,\ldots)$. Then $S$ satisfies condition (2) of Theorem 5 ($s_{n+1}/s_n$ bounded) but not condition (1): $M(S)$ is not complete, since $2\cdot3^n\notin P(M(S))$ for $n=0,1,2,\ldots$. The rational $\tfrac12$ satisfies conditions (3) and (4): it is $(M(S))^{-1}$-accessible, and $2$ divides the term $4$. Yet $\tfrac12\notin P((M(S))^{-1})$.

The paper presents this as showing that condition (1) of Theorem 5 cannot be omitted, and says that no example is known showing the same for condition (2) (p. 205).

Source. R. L. Graham, On finite sums of unit fractions, Proc. London Math. Soc. (3) 14 (1964), no. 2, 193--207, doi:10.1112/plms/s3-14.2.193; the remark after Theorem 5 and the example, pp. 205--206. The edition read is named on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page images of the print, and the argument was checked. Nothing here is independently reviewed.

Proof pointer

Pp. 205--206. Accessibility: $\tfrac12=\sum_{k\ge1}3^{-k}$, and replacing the tail $\sum_{k\ge m+2}3^{-k}=(6\cdot3^m)^{-1}$ by $(4\cdot3^m)^{-1}$ gives a finite subsum equal to $\tfrac12+(12\cdot3^m)^{-1}$ for every $m$. Non-representability: a representation would read $\tfrac12=\sum_i3^{-a_i}+4^{-1}\sum_j3^{-b_j}$ with $a_1\ge1$; multiplying by the largest power of $3$ present and comparing residues modulo $3$ gives a contradiction in each of the cases $a_m<b_n$, $a_m>b_n$ and $a_m=b_n$.

Dependencies

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