利用极限的夹副准则证明limn→无穷大(n/n^2+π+n/n^2+2π+...+n/n^2+nπ)=1
1个回答
展开全部
n/(n^2+nπ) ≤ n/(n^2+mπ) ≤ n/(n^2 + π) 注：n ≤ m ≤ 1
所以,
n*[n/(n^2+nπ)]=n^2/(n^2+nπ) ≤ ∑n/(n^2+mπ) ≤ n*[n/(n^2+π) = n^2/(n^2+π)
因为：lim[n^2/(n^2+nπ)]=lim[1/(1+π/n)] = 1
lim[n^2/(n^2+π)] = lim[1/(1+π/n^2)] = 1
所以,
lim∑n/(n^2+mπ) = 1
所以,
n*[n/(n^2+nπ)]=n^2/(n^2+nπ) ≤ ∑n/(n^2+mπ) ≤ n*[n/(n^2+π) = n^2/(n^2+π)
因为：lim[n^2/(n^2+nπ)]=lim[1/(1+π/n)] = 1
lim[n^2/(n^2+π)] = lim[1/(1+π/n^2)] = 1
所以,
lim∑n/(n^2+mπ) = 1
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
