Question

大神，第5题

Answer 1

(1) 证明
X1 = 2^(1/2) < 2
假设 Xn < 2^(1/2)
有 Xn+1 = ( 2* Xn)^(1/2) < ( 2*2)^(1/2) = 2
由数学归纳法知，
Xn < 2^(1/2)
由于 Xn+1 / Xn = ( 2/Xn)^(1/2) > ( 2/2)^(1/2) = 1
故 有 Xn+1 > Xn
因此 数列 { Xn} 单调增大且有上界，因此，其极限存在，设其为 a ，有
a = (2a)^(1/2)
解得 a = 2

(2) 解：
显然，当 n > 1 时， Xn > 1
X1 = 1 < [ 5^(1/2) + 1 ] / 2
假设 X n < [ 5^(1/2) + 1 ] / 2
Xn+1 = 1 + Xn/(1+ Xn)
= 1 + 1 /( 1/Xn + 1 ) < 1 + 1 / ( 1/ {[ 5^(1/2) + 1 ] / 2 } + 1 ) = [ 5^(1/2) + 1 ] / 2
由数学归纳法知，
X n < [ 5^(1/2) + 1 ] / 2
又 Xn+1 - Xn = 1 + Xn/(1+ Xn) - Xn
= (Xn - [ 5^(1/2) + 1 ] / 2 ) (Xn + [5^(1/2) - 1 ] / 2 ) / (1 + Xn) > 0
故 Xn+1 > Xn
Xn 单调增大且有上界，故 极限存在。
令 lim { Xn} = a
有 a = 1 + a/( 1 + a )
解得 a = [ 5^(1/2) + 1 ] / 2