3. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=2, a_{n}=\frac{1+a_{n-1}}{1-a_{n-1}}(n \geqslant 2) \text {. }
$$

Then the value of $a_{2006}$ is $\qquad$ .
3.2.

Since $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}, a_{5}=$ 2, we conjecture that $a_{n}$ is a periodic sequence with a period of 4. And
$$
a_{n+4}=\frac{1+a_{n+3}}{1-a_{n+3}}=\frac{1+\frac{1+a_{n+2}}{1-a_{n+2}}}{1-\frac{1+a_{n+2}}{1-a_{n+2}}}=-\frac{1}{a_{n+2}}=a_{n} .
$$

Therefore, $a_{2008}=a_{4 \times 501+1}=a_{1}=2$.