4. Calculate $\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{2006}}$.
(Hint: Use the method of borrowing and returning "numbers", that is, add $\frac{1}{2^{2006}}$ to the original expression, then subtract $\frac{1}{2^{2006}}$. Answer: $1-\frac{1}{2^{2006}}$.)