高数微积分能解答下3 4题第二小问怎么写吗？
1个回答
展开全部
(3)
(1)
n/(2n)^2<1/n^2+1/(n+1)^2+...+1/(2n)^2 < n/n^2
lim(n->∞) [n/(2n)^2 ] = lim(n->∞) [n/n^2] =0
=>
lim(n->∞) [ 1/n^2+1/(n+1)^2+...+1/(2n)^2 ] =0
(2)
2^n/n! >0
2^n/n!
= (2/1)(2/2)(2/3)(2/4)...(2/n)
=2(1)(2/3)(2/4)...(2/n)
< 2(2/3)^(n-2)
lim(n->∞) 2(2/3)^(n-2) =0
=> lim(n->∞) 2^n/n! =0
(4)
(1)
xn= 1/(e^n +1)
xn > x(n+1)
|xn|< 1/(0+1) =1
=>
lim(n->∞) 1/(e^n +1) 存在
(2)
x1=√2
x(n+1) = √(2xn)
x(n+1) > xn
x(n+1) = √(2xn)
[x(n+1)]^2 = 2xn
[x(n+1)]^2 - 2xn =0
(xn)^2-2xn <0
xn.(xn-2) <0
0<xn<2
=>
|xn| <2
lim(n->∞) xn 存在
(1)
n/(2n)^2<1/n^2+1/(n+1)^2+...+1/(2n)^2 < n/n^2
lim(n->∞) [n/(2n)^2 ] = lim(n->∞) [n/n^2] =0
=>
lim(n->∞) [ 1/n^2+1/(n+1)^2+...+1/(2n)^2 ] =0
(2)
2^n/n! >0
2^n/n!
= (2/1)(2/2)(2/3)(2/4)...(2/n)
=2(1)(2/3)(2/4)...(2/n)
< 2(2/3)^(n-2)
lim(n->∞) 2(2/3)^(n-2) =0
=> lim(n->∞) 2^n/n! =0
(4)
(1)
xn= 1/(e^n +1)
xn > x(n+1)
|xn|< 1/(0+1) =1
=>
lim(n->∞) 1/(e^n +1) 存在
(2)
x1=√2
x(n+1) = √(2xn)
x(n+1) > xn
x(n+1) = √(2xn)
[x(n+1)]^2 = 2xn
[x(n+1)]^2 - 2xn =0
(xn)^2-2xn <0
xn.(xn-2) <0
0<xn<2
=>
|xn| <2
lim(n->∞) xn 存在
更多追问追答
追问
四题二问不对
xn+1不知是否大于xn
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
