4. Given that $\alpha, \beta$ are the roots of the equation $x^{2}-x-2006=0$. Then the algebraic expression
$$
\alpha+\frac{2006}{1+\frac{2006}{1+\frac{2006}{\beta}}}=(\quad) .
$$
(A) -1
(B) 0
(C) 1
(D) 2006
4. B.

From the relationship between roots and coefficients, and the definition of the roots of the equation, we get $\alpha+\beta=1, \beta^{2}-\beta-2006=0$.
Thus, $\beta+2006=\beta^{2}$.
Dividing both sides by $\beta$ gives $1+\frac{2006}{\beta}=\beta$.
Therefore, the original expression $=\alpha+\frac{2006}{1+\frac{2006}{\beta}}=\alpha+\frac{2006}{\beta}=\alpha+\beta-1=0$.