解:
X1 = 1 < [ 5^(1/2) + 1 ]/2
假设 Xn < [ 5^(1/2) + 1 ]/2
则 Xn+1 = 2 - 1/( 1+Xn) < 2 - 1/ { 1+[ 5^(1/2) + 1 ]/2 } = [ 5^(1/2) + 1 ]/2
由数学归纳法知,
Xn < [ 5^(1/2) + 1 ]/2
又 Xn+1 - Xn = 2 - 1/( 1+Xn) - Xn
= {[ 5^(1/2) + 1 ] /2 - Xn } * {[ 5^(1/2) - 1 ] /2 + Xn } / (1+ Xn) > 0
故 { Xn} 单调增大且有上界。 故其极限存在,并设
lim { Xn } = a
式子 Xn+1 = 2 - 1/( 1+Xn 的两边求极限,有
a = 2 - 1/ (1+a)
解得 a = [ 5^(1/2) + 1 ]/2
2. 解:
因为 (3^n)^(1/n) < an < (3 * 3^n)^(1/n)
即 3 < an < 3 * 3^(1/n)
而 lim { 3 * 3^(1/n) } = 3
故 lim { an } = 3
3. 解:
Xn+1 = (1/2) * ( Xn + a/Xn ) ≥ [ Xn * (a/Xn) ]^(1/2) = a^(1/2)
Xn+1 ≥ a^(1/2)
Xn+1 / Xn = (1/2) * ( Xn + a/Xn ) / Xn
= (1/2) * [ 1 + a/(Xn ) ^2 ] ≥ [ a/(Xn ) ^2 ]^(1/2) = a^(1/2) / Xn ≥ 1
故 { Xn} 单调增大且有上界。 故其极限存在,并设
lim { Xn } = c
式子 Xn+1 = (1/2) * [ 1 + a/(Xn ) ^2 ] 的两边求极限,有
c = (1/2) * [ c + a/c ]
解得 c = a^(1/2)
意思是证明这三个数列收敛并求其极限