$$
\text { 3. } 19 \text { Let } \frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots-\frac{1}{1}}}}=\frac{m}{n} \text {, }
$$

where $m$ and $n$ are coprime natural numbers, and the left side of the equation contains 1988 fraction bars. Calculate the value of $m^{2}+m n-n^{2}$.
[Solution] Let the value of the above complex fraction with $k$ fraction lines be $\frac{m_{k}}{n_{k}}\left(m_{k} 、 n_{k}\right.$ being coprime natural numbers), then
$$
\frac{m_{k+1}}{n_{k+1}}=\frac{1}{1+\frac{m_{k}}{n_{k}}}=\frac{n_{k}}{m_{k}+n_{k}} \text {. }
$$

Notice that $\frac{m_{1}}{n_{1}}=\frac{1}{1}, m_{1}=1, n_{1}=1$;
$$
\begin{aligned}
\frac{m_{2}}{n_{2}} & =\frac{n_{1}}{m_{1}+n_{1}}=\frac{1}{2}, \quad m_{2}=1, n_{2}=2 ; \\
\frac{m_{3}}{n_{3}} & =\frac{n_{2}}{m_{2}+n_{2}}=\frac{2}{1+2}=\frac{2}{3}, \\
m_{3} & =2, n_{3}=3 ;
\end{aligned}
$$
$\qquad$
Therefore, $\frac{m_{k}}{n_{k}}=\frac{n_{k-1}}{m_{k-1}+n_{k-1}}=\frac{F_{k}}{F_{k+1}}$,
where $F_{k}$ is the $k$-th term of the Fibonacci sequence, i.e., $F_{1}=1, F_{2}=1, F_{k+2}=F_{k+1}+$ $F_{k}(k=1,2, \cdots)$
Thus,
$$
\begin{aligned}
& m^{2}+m n-n^{2} \\
= & F_{1988}^{2}+F_{1988} F_{1989}-F_{1989}^{2} \\
= & F_{1988}^{2}+F_{1988}\left(F_{1988}+F_{1987}\right)-F_{1987}^{2} \\
& -2 F_{1988} F_{1987}-F_{1988}^{2} \\
= & -\left[F_{1987}^{2}+F_{1987} F_{1988}-F_{1988}^{2}\right] \\
= & F_{1986}^{2}+F_{1986} F_{1987}-F_{1987}^{2} \\
= & \cdots \cdots \cdots \cdots \\
= & F_{2}^{2}+F_{2} F_{3}-F_{3}^{2} \\
= & 1^{2}+1 \times 2-2^{2} \\
= & -1 .
\end{aligned}
$$