求数列a_n=(1−β)+αβ a_{n−1}的极限问题

求解过程

原题

$$a_k=1-\beta (1-\alpha a_{k-2}),k=2n+1,n\in Z,\alpha ,\beta \in [0,1]$$

转换问题

$$a_n=1-\beta (1-\alpha a_{n-1})=(1-\beta )+\alpha \beta a_{n-1},n\in Z,\alpha ,\beta \in [0,1]$$

解法一

递推：

$$\begin{array}{rl}{a}_n& =(1-\beta )+\alpha \beta a_{n-1}\\ & =(1-\beta )+\alpha \beta ((1-\beta )+\alpha \beta a_{n-2})\\ & =(1-\beta )+\alpha \beta (1-\beta )+{\alpha }^2{\beta }^2{a}_{n-2}\\ & =(1-\beta )+\alpha \beta (1-\beta )+{\alpha }^2{\beta }^2((1-\beta )+\alpha \beta a_{n-3})\\ & =(1-\beta )+\alpha \beta (1-\beta )+{\alpha }^2{\beta }^2(1-\beta )+{\alpha }^3{\beta }^3{a}_{n-3}\\ & =...\\ & =(1-\beta )(1+\alpha \beta +{\alpha }^2{\beta }^2+...+{\alpha }^{n-2}{\beta }^{n-2})+{\alpha }^{n-1}{\beta }^{n-1}\ast a_1\end{array}$$

根据等比数列求和公式，
$$1+\alpha \beta +{\alpha }^2{\beta }^2+...+{\alpha }^{n-2}{\beta }^{n-2}(共有n-1项)=\frac{1-{\alpha }^{n-1}{\beta }^{n-1}}{1-\alpha \beta }$$

所以
$$\begin{array}{rl}{a}_n& =(1-\beta )(1+\alpha \beta +{\alpha }^2{\beta }^2+...+{\alpha }^{n-2}{\beta }^{n-2})+{\alpha }^{n-1}{\beta }^{n-1}\ast a_1\\ & =(1-\beta )\frac{1-{\alpha }^{n-1}{\beta }^{n-1}}{1-\alpha \beta }+{\alpha }^{n-1}{\beta }^{n-1}\end{array}$$

由于$\alpha ,\beta <1$，则
$$\begin{gathered} \underset{n\to \infty }{\lim }{\alpha }^{n-1}{\beta }^{n-1}=0 \\ \underset{n\to \infty }{\lim }(1-{\alpha }^{n-1}{\beta }^{n-1})=1 \end{gathered}$$
因此
$$\underset{n\to \infty }{\lim }{a}_n=\frac{1-\beta }{1-\alpha \beta }$$

解法二

易证该数列的极限存在。设极限为x，则x满足：
$$x=(1-\beta )+\alpha \beta x$$
解得
$$x=\frac{1-\beta }{1-\alpha \beta }$$