如何证明数列X1=2,Xn=2+1/Xn-1在n趋近于无穷时收敛?我想知道证明这个数列的奇子数列是单调递增,偶子数列是单调递减的过程是怎么写的?.

Question

如何证明数列X1=2,Xn=2+1/Xn-1在n趋近于无穷时收敛?
我想知道证明这个数列的奇子数列是单调递增,偶子数列是单调递减的过程是怎么写的?.

Answer 1

x1=2 < x3=12/5 < x4=29/12 < x2=5/2
0 < xn = 2+1/x(n-1)
x(n+2)-xn = 1/x(n+1) - 1/x(n-1) = (x(n-1)-x(n+1))/(x(n-1)x(n+1))
= -1/(x(n+1)x(n-1)) * [x(n+1) - x(n-1)]
= -1/(x(n+1)x(n-1)) * [(x(n-2)-xn)/(x(n-2)xn)]
= 1/[(x(n+1)nx(n-1))(xnx(n-2))] * [(xn-x(n-2)]
∵ 1/[(x(n+1)nx(n-1))(xnx(n-2))] > 0
∴ x(n+2)-xn 与 [(xn-x(n-2)] 同号,即：
x(n-2) < xn ===> xn < x(n+2)
∴x1x6>...>x(2n-2)>x2n
∴ x(2n-1) 与 x2n 为单调有界(上界x2,下界x1）序列,故收敛；由极限保序性,设：
lim x2n = a ≥ x1 = 2
lim x（2n-1） = b ≥ x1
a=lim x2n = lim [2+1/x(2n-1)]=2+1/b
b=lim x(2n+1) = lim [2+1/x2n]=2+1/a
ab=2a+1=2b+1
∴ a=b= 1+√2