3. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}, b_{n}=\frac{1}{1+a_{n}}, S_{n}=b_{1}+b_{2}+\cdots+b_{n}, P_{n}=$ $b_{1} b_{2} \cdots b_{n}$, try to find the value of $2 P_{n}+S_{n}$.
3. Solution: Since $a_{1}=\frac{1}{2}, a_{n-1}=a_{n}^{2}+a_{n}, n \in \mathbf{N}$, it follows that $a_{n-1}>a_{n}>0, a_{n-1}=a_{n}\left(a_{n}+1\right)$

Therefore, $b_{n}=\frac{1}{1+a_{n}}=\frac{a_{n}^{2}}{a_{n} a_{n-1}}=\frac{a_{n-1}-a_{n}}{a_{n} a_{n-1}}=\frac{1}{a_{n}}-\frac{1}{a_{n-1}}, P_{n}=b_{1} b_{2} \cdots b_{n}=\frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdots \frac{a_{n}}{a_{n-1}}=\frac{1}{2 a_{n-1}}$, $S_{n}=b_{1}+b_{2}+\cdots+b_{n}=\left(\frac{1}{a_{1}}-\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{2}}-\frac{1}{a_{3}}\right)+\cdots+\left(\frac{1}{a_{n}}-\frac{1}{a_{n-1}}\right)=2-\frac{1}{a_{n-1}}$,

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