$\because a_{1}=-\dfrac{1}{2}$；
$\therefore a_{2}=\dfrac{1}{1-\left(-\dfrac{1}{2}\right)}=\dfrac{2}{3}$；
$a_{3}=\dfrac{1}{1-\dfrac{2}{3}}=3$，
$\ldots$
$-\dfrac{1}{2}$，$\dfrac{2}{3}$，$3$依次循环，
又$\because 2006\div 3=668\ldots 2$，
$\therefore a_{2006}=\dfrac{2}{3}$.
