$\frac{2}{7} = \frac{1}{5} + \frac{1}{5\cdot 3} + \frac{1}{5\cdot 3\cdot 5} + \frac{1}{5\cdot 3\cdot 5\cdot 3} + \frac{1}{5\cdot 3\cdot 5\cdot 3\cdot 5} + \cdots$

and

$\frac{3}{7} = \frac{1}{3} + \frac{1}{3\cdot 5} + \frac{1}{3\cdot 5\cdot 3} + \frac{1}{3\cdot 5\cdot 3\cdot 5} + \frac{1}{3\cdot 5\cdot 3\cdot 5\cdot 3} + \dots$

More generally,

$\frac{2}{12n+7} = \frac{1}{6n+5} + \frac{1}{(6n+5)(4n+3)} + \frac{1}{(6n+5)(4n+3)(6n+5)} + \cdots$

and

$\frac{3}{12n+7} = \frac{1}{4n+3} + \frac{1}{(4n+3)(6n+5)} + \frac{1}{(4n+3)(6n+5)(4n+3)} + \cdots$

So the analogue to the odd greedy Egyptian fraction problem for Engel expansion has an immediate negative answer.

The same infinite $$2/7$$ and $$3/7$$ expansions also occur as the Costé prime expansion of these numbers, which is infinite for the same reasons.