Three. (Total 20 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, n \in \mathbf{N}, y_{n}=\frac{1}{1+x_{n}}, \\
S_{n}=y_{1}+y_{2}+\cdots+y_{n}, P_{n}=y_{1} y_{2} \cdots y_{n} .
\end{array}
$$

Find $P_{n}+\frac{1}{2} S_{n}$.
$$
\begin{array}{l}
\text { Three, } \because x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, \\
\therefore x_{n+1}>x_{n}>0, x_{n+1}=x_{n}\left(1+x_{n}\right) . \\
\therefore y_{n}=\frac{1}{1+x_{n}}=\frac{x_{n}^{2}}{x_{n} x_{n+1}}=\frac{x_{n+1}-x_{n}}{x_{n} x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}} . \\
\therefore P_{n}=y_{1} y_{2} \cdots y_{n}=\frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}} \cdots \cdots \frac{x_{n}}{x_{n+1}}=\frac{1}{2 x_{n+1}}, \\
S_{n}=\frac{1}{x_{1}}-\frac{1}{x_{2}}+\frac{1}{x_{2}}-\frac{1}{x_{3}}+\cdots+\frac{1}{x_{n}}-\frac{1}{x_{n+1}} \\
\quad=2-\frac{1}{x_{n+1}} .
\end{array}
$$

Therefore, $P_{n}+\frac{1}{2} S_{n}=\frac{1}{2 x_{n+1}}+1-\frac{1}{2 x_{n+1}}=1$ :