14.47 The sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{k+1}=x_{k}^{2}+x_{k}, k \in N$. Find the integer part of the sum $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{100}+1}$.
(China National Training Team Problem, 1990)
[Solution] From $x_{k+1}=x_{k}^{2}+x_{k}$ and $x_{1}=\frac{1}{2}>0$, we can get
$$
x_{k+1}>x_{k}>0 \text {. }
$$

And.
$$
\begin{array}{l}
x_{k+1}-x_{k}^{2}-x_{k}+x_{k} x_{k+1}=x_{k} x_{k+1} \\
\left(x_{k}+1\right)\left(x_{k+1}-x_{k}\right)=x_{k} x_{k+1} \\
\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}} .
\end{array}
$$

Thus, we have
$$
\begin{aligned}
& \frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{100}+1} \\
= & \left(\frac{1}{x_{1}}-\frac{1}{x_{2}}\right)+\left(\frac{1}{x_{2}}-\frac{1}{x_{3}}\right)+\cdots+\left(\frac{1}{x_{100}}-\frac{1}{x_{101}}\right) \\
= & \frac{1}{x_{1}}-\frac{1}{x_{101}} \\
= & 2-\frac{1}{x_{101}} \\
> & 2-\frac{1}{x_{3}} .
\end{aligned}
$$

From $x_{1}=\frac{1}{2}$, we get
$$
x_{2}=\frac{3}{4}, \quad x_{3}=\frac{21}{16}>1 \text {, }
$$

Thus, $\quad 1<\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{100}+1}<2$.
That is, $\left[\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{100}+1}\right]=1$.