请问这里的第三题怎样做?求详细过程
3个回答
展开全部
运用极限的夹逼性求解
因为1/(n^2+1^2)>1/(n^2+2^2)>...>1/(n^2+n^2)
所以n/(n^2+1^2)>1/(n^2+1^2)+1/(n^2+2^2)+...+1/(n^2+n^2)>n/(n^2+n^2)
因为lim(n->∞)n/(n^2+1^2)=lim(n->∞)n/(n^2+n^2)=0
所以根据极限的夹逼性,
lim(n->∞) [1/(n^2+1^2)+1/(n^2+2^2)+...+1/(n^2+n^2)]=0
因为1/(n^2+1^2)>1/(n^2+2^2)>...>1/(n^2+n^2)
所以n/(n^2+1^2)>1/(n^2+1^2)+1/(n^2+2^2)+...+1/(n^2+n^2)>n/(n^2+n^2)
因为lim(n->∞)n/(n^2+1^2)=lim(n->∞)n/(n^2+n^2)=0
所以根据极限的夹逼性,
lim(n->∞) [1/(n^2+1^2)+1/(n^2+2^2)+...+1/(n^2+n^2)]=0
展开全部
lim(n->∞) (1+√2+√3+...+√n ) / (n√n)
=lim(n->∞)(1/n) ∑(i:1->n) √(i/n)
=lim(n->∞)(1/n) ∑(i: 1->n) √(i/n)
=∫(0->1) √x dx
= (2/3)[x^(3/2) ] (0->1)
=2/3
=lim(n->∞)(1/n) ∑(i:1->n) √(i/n)
=lim(n->∞)(1/n) ∑(i: 1->n) √(i/n)
=∫(0->1) √x dx
= (2/3)[x^(3/2) ] (0->1)
=2/3
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