2. Connect the diagonals $AC, BD$ of rectangle $ABCD$ intersecting at point $E$, draw a perpendicular from $E$ to $AB$ intersecting $AB$ at $F$; connect $DF$ intersecting $AC$ at $E_{1}$, draw a perpendicular from $E_{1}$ to $AB$ intersecting $AB$ at $F_{1}$; connect $DF_{1}$ intersecting $AC$ at $E_{2}$, draw a perpendicular from $E_{2}$ to $AB$ intersecting $AB$ at $F_{2}$, and so on. Then $\frac{A F_{2006}}{A B}=$ $\qquad$
2. $\frac{1}{2008}$.

As shown in Figure 6. By the Midline Theorem, we have
$$
\begin{array}{l}
A F=\frac{1}{2} A B, \\
E F=\frac{1}{2} A D .
\end{array}
$$
$$
\text { Also, } \triangle A E_{1} F_{1} \backsim
$$
$\triangle A E F, \triangle A E_{1} D \backsim \triangle E E_{1} F$, thus
$$
\frac{A F_{1}}{F_{1} F}=\frac{A E_{1}}{E_{1} E}=\frac{A D}{E F}=2,
$$

which means $\frac{A F_{1}}{A F_{1}+F_{1} F}=\frac{2}{3}$. Therefore, $A F_{1}=\frac{2}{3} A F=\frac{1}{3} A B$.
Similarly, we have
$$
\begin{array}{l}
A F=\frac{1}{2} A B, A F_{1}=\frac{1}{3} A B, A F_{2}=\frac{1}{4} A B, \\
\cdots \cdots \\
A F_{2006}=\frac{1}{2008} A B .
\end{array}
$$

Thus, $\frac{A F_{2006}}{A B}=\frac{1}{2008}$.