Let $x_{n}$ represent a number in the form of $x$ with $n$ fraction lines. Then, we have $x_{n+1}=\frac{1}{1+x_{n}}$.
It is easy to find $x_{1}=\frac{1}{2}, x_{2}=\frac{2}{3}, x_{3}=\frac{3}{5}, x_{4}=\frac{5}{8}$,
$\cdots$. Let $y_{n}=x_{n}^{2}+x_{n}$, then
$$y_{1}=\frac{3}{4}1, y_{3}=\frac{24}{25}1\$. Conjecture: when \$n\$ is odd, \$y_{n}1\$. We will prove this conjecture.$$
$$\begin{aligned} \because & y_{n+1}-1=x_{n+1}^{2}+x_{n+1}-1 \\ & =\left(\frac{1}{1+x_{n}}\right)^{2}+\left(\frac{1}{1+x_{n}}\right)-1 \\ & =\left(\frac{1}{1+x_{n}}\right)^{2}\left[1+\left(1+x_{n}\right)-\left(1+x_{n}\right)^{2}\right] \\ & =-\left(\frac{1}{1+x_{n}}\right)^{2}\left(x_{n}^{2}+x_{n}-1\right) \\ & =-\left(\frac{1}{1+x_{n}}\right)^{2}\left(y_{n}-1\right), \end{aligned}$$
$$\$\therefore y_{n+1}-1\$ has the opposite sign of \$y_{n}-1\$, which means the conjecture is true. Thus, \$y_{1000}>1\$.$$
$\text{x 2 +x>1}$ {. }
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