Egyptian fractions

Question

Is it true that any fraction $a/b$ can be written in infinitely many ways as a sum of different Egyptian fractions (those with denominator equal to 1)?

Answer 1

The key is the identity: $$\frac1b = \frac{1}{b+1} + \frac{1}{b(b+1)}\tag{*1}$$

For any positive rational number $\frac{a}{b}$, you first express it as a sum of reciprocals of positive integers. $$\frac{a}{b} = \underbrace{\frac{1}{b} + \cdots + \frac{1}{b}}_{a \text{ terms}}$$ Start from any sum of reciprocals, if the corresponding integers are not distinct, pick one of the duplicated integers and apply identity $(*1)$. Repeat this procedure if the remaining integers are not distinct. In finitely many steps, this process will terminate. This end result is an expression of $\frac{a}{b}$ as a sum of reciprocals of distinct positive integers.

$$\frac{a}{b} = \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_{n-1}} + \frac{1}{b_n}$$

where $b = b_1 < b_2 < \cdots < b_n$.

$b_n$ is the "largest integer" appear in above sum. Apply $(*1)$ to this $b_n$, you obtain a new expression of $\frac{a}{b}$ as a sum of distinct reciprocals with larger "largest integer".

$$\frac{a}{b} = \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_{n-1}} + \frac{1}{b_n+1} + \frac{1}{b_n(b_n + 1)}$$

Repeat this procedure, you can generate as many sum of reciprocals for $\frac{a}{b}$ as you wish.