Let (x_n) be a real sequence, and suppose that
|x_(n+2) - x_(n+1)| <= |x_(n+1) - x_n| /√2 for all n in N.
Show that (x_n) is a convergent sequence in R.
高微(收斂數列)
2012-08-26 12:25 am
回答 (1)
2012-08-26 3:27 am
✔ 最佳答案
|x_(n+1)-x_n| ≦ |x_n-x_{n-1}|/√2 ≦ |x_{n-1}-x_{n-2}|/2 ≦ |x_2-x_1|/2^{(n-1)/2}
|x_{n+k}-x_n| ≦ |x_{n+k}-x_{n+k-1}|+...+|x_{n+1}-x_n|
≦ |x_2-x_1|(1/2{(n+k-2)/2}+...+1/2^{(n-1)/2})
≦ (|x_2-x_1|/2^{(n-1)/2})/(1-1/√2)
當 n→∞, k>0, |x_{n+k}-x_n|→0, 即 {x_n} 為一Cauchy sequence.
故 {x_n} 收斂.
收錄日期: 2021-05-04 01:49:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120825000015KK05215