3n
⇒ limｎ→∞ Σ k/(9n^2+k^2)
k=1
救救我，我翻片我手邊的書都找不到類似的題目@@
求極限 sigma
2013-02-01 4:20 am
回答 (2)
2013-02-01 7:56 am
✔ 最佳答案
3nΣ k/(9n^2+k^2) = Σ (1/n)(k/n)/[9+(k/n)^2]
k=1
3
→ ∫ x/(9+x^2) dx = (1/2)ln(9+x^2) ∣_[0,3] = (1/2)ln(18/9)
0
= (1/2)ln(2)
2013-02-01 18:06:39 補充：
k = 1~3n
x_k = k/n = 1/n~3
1/n = △x
2013-02-01 5:33 pm
TO:老大師
Σ (1/n)(k/n)/[9+(k/n)^2]
轉下一步積分的區間如何算出為0至3
還有變為x/(9+x^2) 為何前面係數(1/n)可省略呢?
Σ (1/n)(k/n)/[9+(k/n)^2]
轉下一步積分的區間如何算出為0至3
還有變為x/(9+x^2) 為何前面係數(1/n)可省略呢?
收錄日期: 2021-05-04 01:51:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130131000015KK07486