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证明limn→无穷[(1/n2+n+1)+(2/n2+n+2)+……+(n/n2+n+n)]＝½
题目详情
证明limn→无穷[(1/n2+n+1)+(2/n2+n+2)+……+(n/n2+n+n)]＝½
▼优质解答
答案和解析
当n→无穷,
1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n)
≤1/(n^2+n+1)+2/(n^2+n+1)+...+n/(n^2+n+1)
=(1+2+...+n)/(n^2+n+1)
=(1/2)(n+1)n/(n^2+n+1)
-> 1/2
1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n)
≥1/(n^2+n+n)+2/(n^2+n+n)+...+n/(n^2+n+n)
= (1+2+...+n)/[(n^2)+2n]
= (1/2)(n+1)n/[(n^2)+2n]
-> 1/2
所以,
当n→无穷,1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n)
=1/2
1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n)
≤1/(n^2+n+1)+2/(n^2+n+1)+...+n/(n^2+n+1)
=(1+2+...+n)/(n^2+n+1)
=(1/2)(n+1)n/(n^2+n+1)
-> 1/2
1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n)
≥1/(n^2+n+n)+2/(n^2+n+n)+...+n/(n^2+n+n)
= (1+2+...+n)/[(n^2)+2n]
= (1/2)(n+1)n/[(n^2)+2n]
-> 1/2
所以,
当n→无穷,1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n)
=1/2
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