不动点法,b(n+1)+(1+√2)=1/(bn+2)+(1+√2)=[(1+√2)bn+3+2√2]/(bn+2),
b(n+1)-(√2-1)=1/[b(n)+2]-(√2-1)=[(1-√2)bn+3-2√2]/(bn+2),
两式相除,得
[b(n+1)+(1+√2)]/[b(n+1)-(√2-1)]=[(1+√2)bn+3+2√2]/[(1-√2)bn+3-2√2],
待定系数
得k=[-(1+√2)^2]
即b(n+1)=1/(bn+2)可化为
[b(n+1)+(1+√2)]/[b(n+1)-(√2-1)]=[-(1+√2)^2]*[bn+(1+√2)]/[bn-(√2-1)]
即Cn=[bn+(1+√2)]/[bn-(√2-1)]是以[-(1+√2)^2]为公比的等比数列.