高微(convergence of {|s_n|})

👨🏻‍🏫 學術台數學

用戶40851

2012-09-19 3:35 am

Let {s_n} be a sequence of real numbers. Show that the convergence of {s_n}
implies the convergence of {|s_n|}. Does the converse hold ?

更新1:

謝謝老怪物的回答，但是does converse hold 我網路上查到的好像是False, 他舉的例子是s_n={(-1)^n}, |s_n|=1, s_n not convergent

回答 (6)

老怪物

2012-09-19 5:13 am

✔ 最佳答案

設 s_n→s 當 n→∞.
則, for any ε>0 there exists N such that
whenever n>N then |s_n-s|<ε.

由
||s_n|-|s|| ≦ |s_n-s|

因此, whenever n>N we have ||s_n|-|s|| < ε.

By definition, {|s_n|} converges to |s| provided that
{s_n} converges to s.

2012-09-19 10:12:30 補充：
{|s_n|} 收斂確實不 implies {s_n} 收斂.

👍 0 👎 0

[匿名]

2014-06-03 2:18 pm

到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

👍 0 👎 0

[匿名]

2014-05-28 3:51 pm

參考下面的網址看看

http://phi008780520.pixnet.net/blog

👍 0 👎 0

[匿名]

2014-05-24 1:27 pm

參考下面的網址看看

http://phi008780520.pixnet.net/blog

👍 0 👎 0

[匿名]

2014-05-22 11:52 pm

參考下面的網址看看

http://phi008780520.pixnet.net/blog

👍 0 👎 0

[匿名]

2012-09-19 5:35 am

converse當然不成立
如sn=1,-1,1,-1..
則|sn|->1但sn不存在

👍 0 👎 0

收錄日期: 2021-05-04 01:48:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120918000010KK06153

檢視 Wayback Machine 備份