Convergence 2(20points)

2007-03-11 10:32 am
a) Given that (∞Σn=1)an converges and an≧0 for any n belongs to Natural numbers.
i) Prove that there exists N(which is belongs to Natural numbers) such that for any n (Natural numbers) if n >N then 0≦an^2≦an. 20 points
ii) Prove that (∞Σn=1)an^2 converges.


b) If (∞Σn=1)an^2 converges is it necessary true that (∞Σn=1)an converges?
Give reason to justify your assertion.

只需答part b,a part已解~
更新1:

問:你係 part a solved 還是 part b solved? 我只看此問題所以以為 part a solved,但係以後睇番另一題,你係問 part b wor. 覆:這一條題目有part a 和 b,我已經識做part a ,在這裡只是想請教各位part b點做,故答part b 已經足夠(post埋part a出來是因為part b 可能要用part a results,thz)

回答 (2)

2007-03-13 7:45 pm
✔ 最佳答案
Please find the proof for the counter example of an = 1/n as follows:

圖片參考:http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Sequencelim13.jpg


圖片參考:http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Sequencelim14.jpg
參考: My Maths knowledge
2007-03-11 10:46 am
b)
Answer is no.

Let an be 1/n, which is 1/n>0 for all natural number n,
Consider the integration of 1/n^2 (from 1 to infinity and from 2 to infinity),
it is obvious that (∞Σn=1)1/n^2 converges.

However, (∞Σn=1)1/n does not converges.
If it does converge to some number, let say k is an integer greater than that convergence, we have
((2^*(k+1))Σn=1)1/n > k
contradication.


2007-03-11 02:49:10 補充:
你係 part a solved 還是 part b solved? 我只看此問題所以以為 part a solved,但係以後睇番另一題,你係問 part b wor.


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