19. (15 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{n+1} a_{n}+3 a_{n+1}+a_{n}+4=0 \text {. }
$$

If $a_{2006}$ is the minimum term of the sequence $\left\{a_{n}\right\}$, find the range of the first term $a_{1}$.
19. Solution 1: From the problem condition, we get $a_{n+1}=\frac{-a_{n}-4}{a_{n}+3}$.

Let $a_{n+1}-\lambda=\frac{\alpha\left(a_{n}-\lambda\right)}{a_{n}+3}$, i.e.,
$$
a_{n+1}=\frac{(\alpha+\lambda) a_{n}-\alpha \lambda+3 \lambda}{a_{n}+3} .
$$

By comparing coefficients, we get
$$
\left\{\begin{array}{l}
\alpha+\lambda=-1, \\
-\alpha \lambda+3 \lambda=-4 .
\end{array}\right.
$$

Solving these, we get $\alpha=1, \lambda=-2$.
Thus, $a_{n+1}+2=\frac{a_{n}+2}{a_{n}+3}$.
Therefore, $\frac{1}{a_{n+1}+2}=\frac{1}{a_{n}+2}+1$
$$
=\frac{1}{a_{n-1}+2}+2=\cdots=\frac{1}{a_{1}+2}+n \text {. }
$$

So, $a_{n}=-2+\frac{1}{n-\left(1-\frac{1}{2+a_{1}}\right)}$.
It is easy to see that the function $f(n)=\frac{1}{n-a}\left(n \in \mathbf{N}_{+}\right)$ reaches its minimum value when $n-a>0$. Therefore, since $a_{2006}$ is the minimum term of $\left\{a_{n}\right\}$, we have
$$
20060$ when, the sequence $\left\{\frac{1}{a_{n}+2}\right\}$ is a positive increasing sequence, and there is no maximum term in this sequence, thus, there is no minimum term in the sequence $\left\{a_{n}\right\}$, which contradicts the fact that $a_{2} \cos$ is the minimum term of $\left\{a_{n}\right\}$. Hence $\frac{1}{a_{1}+2}0$.
So, $\frac{1}{a_{2000}+2}=\frac{1}{a_{2007}+2}-1>-1$.
Thus, $-1<\frac{1}{a_{2006}+2}<0$, i.e.,
$-1<\frac{1}{a_{1}+2}+2005<0$.
Solving this, we get $-\frac{4011}{2005}<a_{1}<-\frac{4013}{2006}$.