求limit值(2)

2010-02-08 11:56 pm
已知0<a(1)<1, a(n+1)=arctan[a(n)], n>0, 則 lim(n->∞) (√n * a(n) )=?
更新1:

Ans: √(3/2)

回答 (4)

2010-02-21 6:32 am
✔ 最佳答案
性質: lim(n->∞)[a(n+1)-a(n)]=A, 則lim(n->∞) a(n)/n=A
利用以上性質比較簡單!

2010-02-20 22:32:34 補充:
1. 顯然 a(n)遞減至0 (至少n夠大時如此)
2. 設b(n)= 1/[a(n)]^2, 則
lim(n->∞) [b(n+1)-b(n)]
=lim(n->∞)[ 1/arctan^2(a(n)) - 1/a(n)^2]
(代換x=arctan(a(n)) 得 lim(x->0) (1/x^2 - 1/tan^2(x))
= lim(x->0)[ (tanx +x)(tanx - x)/(xtanx)^2]
= 2/3 (以Taylor's expansion計算,請自行計算, OK!?)
3. 由意見所述性質知 lim(n->∞) b(n)/n = 2/3
則 lim(n->∞) n*[a(n)]^2 = 3/2
故原題limit =√(3/2)
2010-02-12 12:40 am
分析能力很強,嚴謹度可再補強嗎?

2010-02-20 22:13:46 補充:
有點籠統,請學長講仔細一點,好嗎?

2010-02-20 22:33:59 補充:
謝 謝!
2010-02-10 4:00 am
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2010-02-09 20:00:24 補充:
http://img163.imageshack.us/img163/1293/79790581.png
2010-02-09 5:44 am
call 高錕啦
佢真係可以幫到你嫁
唔啱call 陳振聰啦
佢乜都識嫁
山頂可賦山道 24號


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