Question

设x1>0,Xn+1=1/2（Xn+1/Xn）（n=1,2,3....n），证明数列极限Xn n趋向无穷存在 并且求极限值

好的加分

Answer 1

如下

Answer 2

X1>0, Xn+1 = 1/2 (Xn +1/Xn) ≥ 1
Xn+1 - 1/2 Xn = (1/2) / Xn ≤ 1/2
Xn - (1/2) Xn-1 ≤ 1/2 (1)
=> (1/2) Xn-1 - (1/2^2) Xn-2 ≤ 1/ 2^2 (2)
......
[1/2^(n-2)) X2 - [1/2^(n-1)] X1 ≤ 1/ 2^n (n)
上面n个式子相加：Xn - [1/2^(n-1)] X1 ≤ 1/2 + 1/2^2 + ...... + 1/2^n
当n->∞时, [1/2^(n-1)] X1 -> 0 , 1/2 + 1/2^2 + ...... + 1/2^n -> 1
于是 lim Xn = 1

Answer 3

x=(x+1/x)/2 ==> x^2-2x-1=0==>x=1
n->inf xn-->1
xn>1 x_(n+1)<(x_n+1)/2,
x_n=(x_n+1)/2 收敛，故收敛x_(n+1)<(x_n+1/x_n)/2。