convergence(urgent！)

我估貓朋有些地方做錯了。
a) (∞Σn=1)√ {(3n+1 )/(16n-3)}
Consider lim(n→∞) {(3n+1 )/(16n-3)}
lim(n→∞) {(3n+1 )/(16n-3)}
=lim(n→∞) {(3+1/n )/(16-3/n)}
=3/16
so lim(n→∞)√ {(3n+1 )/(16n-3)}=√(3/16)
By zero test, since lim(n→∞)an not equal to 0
(∞Σn=1)√ {(3n+1 )/(16n-3)} is divergent.
(b)
{(3n+1 )/(16n^2 -3)}
>= {(3n+1 )/(16n^2)}
>= {(3n)/(16n^2)}
=3/16n
Consider the sequence an= 3/16n and (∞Σn=1)an
Since (∞Σn=1)1/n is divergent
So (∞Σn=1)an is divergent
By comparsion test
(∞Σn=1)√ {(3n+1 )/(16n^2 -3)} is also divergent








2007-03-12 13:56:48 補充：
個(b)漏了根號﹐不過an全部都是正數﹐所以其實冇分別{(3n+1 )/(16n^2 -3)}>= {(3n+1 )/(16n^2)}>= {(3n)/(16n^2)}=3/16nSimilarly √ {(3n+1 )/(16n^2 -3)} >=√ (3/16n)

2007-03-12 13:57:09 補充：
Consider the sequence an= √ (3/16n) and (∞Σn=1)anSince (∞Σn=1)√ (1/n) is divergentSo (∞Σn=1)an is divergentBy comparsion test (∞Σn=1)√ {(3n+1 )/(16n^2 -3)} is also divergent

2007-03-12 13:58:52 補充：
(∞Σn=1)√ (1/n) is divergent 可以在任何分析書找到可以當條定理背了

a) (3n+1 )/(16n-3) ≦ (3n+n )/(16n-3) < (4n )/(16n) = 1/4n < 1/n


(∞Σn=1)√ {1/n} diverges by integral
Hence,
(∞Σn=1)√ {(3n+1 )/(16n-3)} is divergent by comparion test

b) (3n+1 )/(16n^2 -3) ≦ (3n+1 )/(16n-3)

Henc, (∞Σn=1)√ {(3n+1 )/(16n^2 -3)} is divergent by comparion test and part (a)



圖片參考：http://i175.photobucket.com/albums/w130/bjoechan2003/My%20Cat%2020070228/DSCN0687.jpg?t=1172589743


2007-03-11 10:28:10 補充：
a) (∞Σn=1)√ {1/n} diverges by integral test

2007-03-12 20:56:55 補充：
哎呀，邪門，邪門，已經第二題，哩張貓相真係唔吉利，竟然將個 inequality 掉能轉做仲懵剩剩。真係抵比人吋係豬犀。謝謝 myisland8132 指證。雖然無可能被選，不過我唔會就咁取消答案，而且我一定唔信邪，會繼續用哩張貓相答數學題。

2007-03-12 21:02:30 補充：
(3n 1 )/(16n-3) (3n)/(16n) = 3/16 1/16(∞Σn=1)√ {1/16} = (∞Σn=1){1/4} divergesHence,(∞Σn=1)√ {(3n 1 )/(16n-3)} is divergent by comparison test

2007-03-12 21:06:57 補充：
(3n+1 )/(16n^2 -3) &gt; (3n)/(16n^2) = 3/16n &gt; 1/16n(∞Σn=1)√ {1/16n} diverges by integral testHence, (∞Σn=1)√ {(3n+1 )/(16n^2 -3)} is divergent by comparison test