高微(收斂數列)

2012-08-26 12:25 am
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.

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.


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