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

#NumberTheory #Limits #InfiniteSeries #EgyptianFractions

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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

Maggie Miller - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$