[数列] 不动点数列

mathworld

题目：  设$f_1 ( x ) =\frac {2}{x+1}$, 定义$f_{n+1} ( x ) = f_1 ( f_n ( x ) )$,

           $a_n = \frac {f_{n}( 0 ) - 1}{f_{n}( 0 ) - 2},n >1$

          求$a_n$的通项公式.

mathworld

设$b_n=f_n(0), b_1 =2,  b_{n+1}=f_1( b_n ) =\frac {2}{1+b_n}$


$$b_{n+1}+2=\frac {2b_n+4}{1+b_n}$$

$$b_{n+1}-1=\frac {1-b_n}{1+b_n}$$

$$\frac {b_{n+1}-1}{b_{n+1}+2} = - \frac 12\frac {b_{n}-1}{b_{n}+2} $$下略