Let the "Infinite Egyptian Expansion" of a real number 0 < r < 1 be defined as follows. It is an infinite series of fractions \(\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{a_n}\), where \(a_n \in \mathbb{N}\), and for each \(n\) we have \(a_n\) as small as possible such that the \(n^{th}\) partial sum is (strictly) less than \(r\).

For example, if \(r = \dfrac{1}{2}\), the Infinite Egyptian Expansion would be \(\dfrac{1}{3}+\dfrac{1}{7} + \dfrac{1}{43} + \dfrac{1}{1807}+\dfrac{1}{3263443}+...\)

Prove that for all \(r \in \mathbb{Q}\), we have \[\lim_{n \to \infty} \frac{a_{n+1}}{a_n^2} = 1\]

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TopNewestLet \(\frac{p}{q}\in(0,1)\), and \(a_n\) as above. Suppose \(p>1\). Note \(\displaystyle\frac{p}{q}-\frac{1}{a_1}=\frac{pa_1-q}{qa_1}\), and

\(\displaystyle\frac{p}{q}<\frac{1}{a_1-1}\) (using \(p>1\) to get strict inequality)

\(\displaystyle pa_1-q<p\).

Therefore, for some \(n,m\), \(\displaystyle\frac{p}{q}-\sum_{k=1}^na_k=\frac{1}{m}\).

Since we are only interested in the limit behavior of the \(a_n\), we may assume \(p=1\).

Then from the identity \(\displaystyle\frac{1}{m}=\frac{1}{m+1}+\frac{1}{m(m+1)}\), we find

\(\displaystyle a_1=q+1, a_{n+1}=a_n(a_n-1)+1\).

Therefore, \(\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{{a_n}^2}=\lim_{n\to\infty}\frac{n(n-1)+1}{n^2}=1\). – Maggie Miller · 1 year, 9 months ago

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