4. Let $f_{1}(x)=\frac{2}{1+x}$. Define $f_{n+1}(x)=f_{1}\left[f_{n}(x)\right]$. And let $a_{n}=\frac{f_{n}(0)-1}{f_{n}(0)+2}$. Then $a_{100}=$ $\qquad$ .
4. $-\frac{1}{2^{101}}$.

Since $f_{1}(0)=2$, we have $a_{1}=\frac{1}{4}$.
From $f_{1}(x)=\frac{2}{1+x}$, we get $f_{n}(0)=\frac{2}{1+f_{n-1}(0)}$.
Thus, $a_{n}=\frac{f_{n}(0)-1}{f_{n}(0)+2}=-\frac{1}{2} \frac{f_{n-1}(0)-1}{f_{n-1}(0)+2}$,
which means $a_{n}=-\frac{1}{2} a_{n-1}$.
Therefore, $a_{n}=\frac{1}{4}\left(-\frac{1}{2}\right)^{n-1}, a_{100}=-\frac{1}{2^{101}}$.