10. (20 points) Given the sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=\frac{1}{3}, x_{n+1}=x_{n}^{2}+x_{n}(n=1,2, \cdots) \text {. }
$$

Find the value of $\sum_{n=1}^{2002} \frac{1}{x_{n}+1}$ between which two integers.
10. Notice that $x_{n}^{2}+x_{n}=x_{n}\left(x_{n}+1\right)$.
$$
\begin{array}{l}
\text { Then } \frac{1}{x_{n}\left(x_{n}+1\right)}=\frac{1}{x_{n}}-\frac{1}{x_{n}+1} \\
\Rightarrow \frac{1}{x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{x_{n}+1} \\
\Rightarrow \frac{1}{x_{n}+1}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}} . \\
\text { Hence } \sum_{n=1}^{200} \frac{1}{x_{n}+1}=\frac{1}{x_{1}}-\frac{1}{x_{2003}}=3-\frac{1}{x_{2003}} . \\
\text { Also } x_{2}=\frac{4}{9}, x_{3}=\frac{52}{81}, x_{4}=\frac{6916}{6561}>1, \text { and the sequence }
\end{array}
$$

$\left\{x_{n}\right\}$ is monotonically increasing, so
$$
\sum_{n=1}^{2002} \frac{1}{x_{n}+1} \in(2,3) \text {. }
$$