# Infinite Egyptian Fractions

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 $\lim_{n \to \infty} \frac{a_{n+1}}{a_n^2} = 1$

Note by Ariel Gershon
2 years, 10 months ago

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Let $$\frac{p}{q}\in(0,1)$$, 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, $$\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{{a_n}^2}=\lim_{n\to\infty}\frac{n(n-1)+1}{n^2}=1$$.

- 2 years, 10 months ago