Representation

The following greedy algorithm can represent any $0\le \frac{a}{b} <1$ as such a sum:

1. Let $$ n = \left \lceil \frac{b}{a} \right \rceil $$ be the smallest natural number for which $\frac{1}{n} \le \frac{a}{b}$ If $a=0$ terminate.
2. Output $\frac{1}{n}$ as the next term of the sum.
3. Continue from step 1, setting $$\frac{a'}{b'} = \frac{a}{b}-\frac{1}{n}.$$

Proof of correctness

Proof. The algorithm can never output the same unit fraction twice. Indeed, any $n$ selected in step 1 is at least 2, so $\frac{1}{n-1} < \frac{2}{n}$ - so the same $n$ cannot be selected twice by the algorithm, as then $n-1$ could have been selected instead of $n$

It remains to prove that the algorithm terminates. We do this by induction on $a$

For $a=0$

The algorithm terminates immediately.

For $a>0$

The $n$ selected in step 1 satisfies $$b \le an < b+a.$$ So $$ \frac{a}{b}-\frac{1}{n} = \frac{an-b}{bn}, $$ and $0 \le an-b < a$ - by the induction hypothesis, the algorithm terminates for $\frac{a}{b}-\frac{1}{n}$

Problems

1. The greedy algorithm always works, but it tends to produce unnecessarily large denominators. For instance, $$\frac{47}{60}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5},$$ but the greedy algorithm selects $\frac{1}{2}$ leading to the representation $$ \frac{47}{60}=\frac{1}{2}+\frac{1}{4}+\frac{1}{30}. $$
2. The representation is never unique. For instance, for any $n$ we have the representations $$ \frac{1}{n} = \frac{1}{n+1} + \frac{1}{n\cdot(n+1)} $$ So given any one representation of $\frac{a}{b}$ as a sum of different unit fractions we can take the largest denominator appearing $n$ and replace it with two (larger) denominators. Continuing the process indefinitely, we see infinitely many such representations, always.