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Source. Theorem 5, arXiv:2202.00191v2, PDF p. 9 (Section 4); proof pp. 9--10, using Theorem 1 (p. 3) and Theorem 4 (p. 8, the Muirhead-type inequality proved in the Appendix, pp. 18--20). Published as J. Number Theory 242 (2023), 208--234; not compared.

Statement

An $n$-term Egyptian underapproximation sequence of $\theta\in(0,1]$ is a sequence of integers $2\le x_1\le x_2\le\cdots\le x_n$ (repetitions allowed) with $\sum_{i\le n}1/x_i<\theta$; the greedy sequence $(a_i)$ of $\theta$ has $a_1=G(\theta)=\lfloor1/\theta\rfloor+1$ and $a_{i+1}=G(\theta-\sum_{j\le i}1/a_j)$.

Theorem 5. Let $p$ and $q$ be positive integers with $p\mid q+1$ and $\theta=p/q\le1$, and let $(a_i)_{i\ge1}$ be the greedy sequence of $\theta$. Fix $n\ge1$. If an $n$-term Egyptian underapproximation sequence $(x_i)_{i\le n}$ of $\theta$ has reciprocal sum at least the greedy one,

$$\sum_{i=1}^n\frac1{a_i}\le\sum_{i=1}^n\frac1{x_i}<\frac pq,$$

then it is the greedy sequence: $x_i=a_i$ for $i=1,\ldots,n$.

So the greedy $n$-term sum is the unique best $n$-term underapproximation of $p/q$, for every $n$. Since $(a_i)$ is strictly increasing ($a_{i+1}\ge a_i^2-a_i+1$), the same holds among distinct denominators. Theorem 1 (p. 3) gives the sequence explicitly: $a_1=(q+1)/p$, $a_{k+1}=qa_1\cdots a_k+1$, and $p/q-\sum_{i\le k}1/a_i=1/(qa_1\cdots a_k)$; for $\theta=1$ this is Sylvester's sequence $2,3,7,43,\ldots$ (Corollary 1).

Proof structure (pp. 9--10)

Induction on $n$. The hypothesis and Theorem 1 give $0<p/q-\sum1/x_i\le1/(qa_1\cdots a_n)$, while $p/q-\sum1/x_i$ is a positive multiple of $1/(qx_1\cdots x_n)$, so $\prod a_i\le\prod x_i$. Let $m\le n-1$ be the largest index with $\prod_{i>m}a_i\le\prod_{i>m}x_i$; maximality gives $\prod_{i=m+1}^{m+j}a_i\le\prod_{i=m+1}^{m+j}x_i$ for all $j\le n-m-1$. If $(x_i)_{i>m}\ne(a_i)_{i>m}$, Theorem 4 (the Muirhead-type inequality on which Soundararajan's method rests: for increasing sequences with those product inequalities, $\sum_{i>m}1/x_i<\sum_{i>m}1/a_i$) yields a contradiction with the hypothesis; hence $x_i=a_i$ for $i>m$, and the induction hypothesis applies to the first $m$ terms.

Read depth

Claims checked (statement read clause by clause on PDF p. 9); the proof was read for structure; not rewritten in full and not independently reviewed.

Bears on. #206: the case $a\mid b+1$ with $a/b\le1$ of the site's commentary; Chu's Theorem 1.12 extends it.