  Let $\mathcal{G}$ be the greedy algorithm that, for each $\theta\in (0,1]$,
produces an infinite sequence of positive integers $(a_n)_{n=1}^\infty$
satisfying $\sum_{n=1}^\infty 1/a_n = \theta$. For natural numbers $p < q$, let
$\Upsilon(p,q)$ denote the smallest positive integer $j$ such that $p$ divides
$q+j$. Continuing Nathanson's study of two-term underapproximations, we show
that whenever $\Upsilon(p,q) \leqslant 3$, $\mathcal{G}$ gives the (unique)
best two-term underapproximation of $p/q$; i.e., if $1/x_1 + 1/x_2 < p/q$ for
some $x_1, x_2\in \mathbb{N}$, then $1/x_1 + 1/x_2 \leqslant 1/a_1+1/a_2$.
However, the same conclusion fails for every $\Upsilon(p,q)\geqslant 4$.
  Next, we study stepwise underapproximation by $\mathcal{G}$. Let $e_{m} =
\theta - \sum_{n=1}^{m}1/a_n$ be the $m$th error term. We compare $1/a_m$ to a
superior underapproximation of $e_{m-1}$, denoted by $N/b_m$ ($N
\in\mathbb{N}_{\geqslant 2}$), and characterize when $1/a_m = N/b_m$. One
characterization is $a_{m+1} \geqslant N a_m^2 - a_m + 1$. Hence, for rational
$\theta$, we only have $1/a_m = N/b_m$ for finitely many $m$. However, there
are irrational numbers such that $1/a_m = N/b_m$ for all $m$. Along the way,
various auxiliary results are encountered.
