Question

高中数学求极限

lim n^2{100/n-[1/(n+1)+1/(n+2)+……1/(n+100)]}问一下这个式子答案为多少？怎么求
n=∞
谢谢

Answer 1

用罗必塔法则先放缩
设A=1/(n+1)+1/(n+2)+……+1/(n+100)
=[1/(n+1)+1/(n+100)]+[1/(n+2)+1/(n+99)]+……+[1/(n+50)+1/(n+51)](共50组）
则50[1/(n+50)+1/(n+51)]<A<50[1/(n+1)+1/(n+100)]
即(50*101n+50*51*100)/n(n+50(n+51)<100/n-A<(50*101n+100*100)[1/n(n+1)(n+100)
lim n^2(50*101n+50*51*100)/n(n+50(n+51)=50*101=5050
lim n^2(50*101n+100*100)[1/n(n+1)(n+100)=50*101=5050
lim n^2{100/n-[1/(n+1)+1/(n+2)+……1/(n+100)]}
=lim n^2(100/n-A)=5050

以下是错误的放缩。两边的极限要是一样的，那中间的极限不言自明

100/(n+100)<1/(n+1)+1/(n+2)+……+1/(n+100)<100/(n+1)
100/n-100/(n+100)<100/n-[1/(n+1)+1/(n+2)+……1/(n+100)]<100/n-100/(n+1)
10000/(n^2+100n)<100/n-[1/(n+1)+1/(n+2)+……1/(n+100)]<100/(n^2+n)
lim n^2*10000/(n^2+100n)=10000
lim n^2*100/(n^2+n)=100

追问

问一下那个 则 那一句是如何推导到 即 那一句的？