3.3.5 ** The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}, n \in \mathbf{N}^{*}, b_{n}=\frac{1}{1+a_{n}}$,
$$
S_{n}=b_{1}+b_{2}+\cdots+b_{n}, P_{n}=b_{1} b_{2} \cdots b_{n},
$$

Find the value of $2 P_{n}+S_{n}$.
Given that $a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}, n \in \mathbf{N}^{*}$, therefore, $a_{n+1}=a_{n}\left(a_{n}+1\right)$, thus
$$
\begin{array}{c}
b_{n}=\frac{1}{1+a_{n}}=\frac{a_{n}^{2}}{a_{n} a_{n+1}}=\frac{a_{n+1}-a_{n}}{a_{n} a_{n+1}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\
P_{n}=b_{1} b_{2} \cdots b_{n}=\frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdots \cdot \frac{a_{n}}{a_{n+1}}=\frac{1}{2 a_{n+1}} \\
S_{n}=b_{1}+b_{2}+\cdots+b_{n} \\
=\left(\frac{1}{a_{1}}-\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{2}}-\frac{1}{a_{3}}\right)+\cdots+\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right)=2-\frac{1}{a_{n+1}}
\end{array}
$$

Therefore, $2 P_{n}+S_{n}=\frac{1}{a_{n+1}}+\left(2-\frac{1}{a_{n+1}}\right)=2$.