作者Dirichlet ( )
看板Math
标题Re: [分析] 初微(28)
时间Thu Aug 11 03:50:59 2005
※ 引述《PttFund (批踢踢基金只进不出)》之铭言：
: Suppose that |a_n| < 2 and
: |a_(n+2) - a_(n+1)|≦ (1/8)| ( a_(n+1) )^2 - ( a_n )^2 |
: for all positive integers n. Show that {a_n} converges.
|a_(n+2) - a_(n+1)|
≦ (1/8) | ( a_(n+1) )^2 - ( a_n )^2 |
= (1/8) | (a_(n+1)) + a_n| |a_(n+1) - a_n|
< (1/2) |a_(n+1) - a_n| for every n≧1
< (1/4) |a_n - a_(n-1)| < ...
< (1/2^n) |a_2 - a_1|
< 4/2^n = 1/2^(n-2)
Claim : {a_n} 是 Cauchy seq.
For any m > n =>
|a_m - a_n|
≦ |a_m - a_(m-1)| + ... + |a_(n+1) - a_n|
< 1/2^(m-4) + ... + 1/2^(n-3)
1/2^(n-3)[1-(1/2)^(m-n)]
= ------------------------
1/2
< 1/2^(n-4) -> 0 as n -> oo
故 {a_n} 是 Cauchy seq. 且必收敛
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