早教吧作业答案频道 -->其他-->
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
看了 证明limn→无穷[(1/n...的网友还看了以下:
定义映射f:A→B,其中A={(m,n)|m,n∈R}接着 B=R,已知对所有的有序正整数对(m, 2020-05-16 …
判断命题是正确与否1、α∥β,m∈α则m∥β2,、m∥α,n∈α则m平行n3.α⊥β,m∥α,则m 2020-05-20 …
濡傛灉m-3n+4=0,闾d箞(m-3n)虏+7m鲁-3(2m鲁n-m虏n-1)+3(m鲁+2m鲁 2020-07-01 …
不等式的证明设m,n为正整数,f(n)=1+1/2+1/3+.+1/n,证明(1)若n>m,则f( 2020-07-16 …
1.如果不等式x>a+2,x<a-3无解,试判断x>2-a,x<a+2的情况2.如果x≥m,x≤n 2020-07-31 …
一天,老师布置了一份课外作业,在由m×n(m×n>1)个小正方形组成的正方形网格中,当m、n互质( 2020-07-31 …
已知m+n=1,mn=-1/2,利用因式分解(提公因式法),求m(m+n)(m-n)-m(m+n) 2020-08-03 …
如图,在△ABC中,AD是∠A的外角平分线,P是AD上异于A的任意一点,设PB=m,PC=n,AB 2020-08-03 …
设数列an满足a1=2,a(m+n)+a(m-n)-m+n=1/2(a2m+a2n)..设数列an满 2020-10-31 …
关于M^n/n!(M>0为常数)求当n趋向于正无穷时的极限,我用斯特林公式将n!代换为什么无法求出极 2021-01-07 …