lim(1/(n+1)+1/(n+2)+1/(n+3)+…1/(n+n))
n->無限
請問這題微積分怎麼解?
2016-04-16 1:21 pm
回答 (2)
2016-04-16 10:36 pm
✔ 最佳答案
1/(n+1)+1/(n+2)+1/(n+3)+…1/(n+n))=(1/n)[1/(1+1/n)+1/(1+2/n)+1/(1+3/n)+…1/(1+n/n))]右式就是估算定積分 "(1/x) dx from x=1 to x=2" 之 黎曼和(Riemann sum)，故極限為該定積分=ln2.
2016-04-17 11:02 am
lim(1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+n))
n->無限
=lim (1/n)/(1+1/n) +(1/n)/(1+2/n) +(1/n)/(1+3/n)+...+(1/n)/(1+1)
n->無限
=0
n->無限
=lim (1/n)/(1+1/n) +(1/n)/(1+2/n) +(1/n)/(1+3/n)+...+(1/n)/(1+1)
n->無限
=0
收錄日期: 2021-04-18 14:46:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160416052105AAaLIc9