作者Dirichlet ( )
看板Math
标题Re: [分析] 初微(49)
时间Thu Aug 11 03:49:43 2005
※ 引述《TaiwanBank (澳仔金控台湾分行)》之铭言：
: If a_1 = 1 and a_(n+1) = (1 + a_n)^(1/2) for n = 1, 2,...
: Show that {a_n} is bounded and converges. What is the limit?
a_1 = 1 < 2, 设 a_n < 2
a_(n+1) = (1 + a_n)^(1/2) < √3 < 2
归纳法 => a_n < 2 for every n≧1
故 {a_n} 有上界
(a_1 = 1 > 0, 类似地归纳法可得 {a_n} 有下界.)
a_2 = √(1+1) = √2 => a_2 - a_1 > 0, 设 a_n - a_(n-1) > 0
a_(n+1) - a_n = (1 + a_n)^(1/2) - a_n > [1 + a_(n-1)]^(1/2) - a_n
= a_n - a_n = 0
归纳法 => a_(n+1) > a_n for every n≧1
故 {a_n} 递增
{a_n} 是递增有上界数列 => {a_n} 是收敛数列
设 a_n -> x as n -> oo => lim a_(n+1) = lim (1 + a_n)^(1/2)
n->oo n->oo
x = (1+x)^1/2 => x^2 - x - 1 = 0 => x = (1±√5)/2 (负不合)
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