lim(1/(n^2+4)^1/2+...+1/(n^2+4n^2)^1/2) n->无穷

来源：学生作业帮助网编辑：作业帮时间：2024/05/11 12:30:29

lim(1/(n^2+4)^1/2+...+1/(n^2+4n^2)^1/2) n->无穷
lim(1/(n^2+4)^1/2+...+1/(n^2+4n^2)^1/2) n->无穷

lim(1/(n^2+4)^1/2+...+1/(n^2+4n^2)^1/2) n->无穷
1/(n^2+4)^1/2+...+1/(n^2+4n^2)^1/2)
=∑1/n{1/√[1+4(k/n)^2]}
所以,根据导数定义
原极限=lim∑1/n{1/√[1+4(k/n)^2]}=∫1/√[1+4x^2]dx 积分范围0到1
令x=1/2tana,a属于[0,arctan2]
∫（0到1）1/√[1+4x^2]dx
=1/2∫（0到arctan2）secada
=1/2 ln|seca+tana| | 0到arctan2
=(1/2)ln(2+√5)

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