已知n属于N,n大于等于2,证明:1/2小于1/n+1+1/n+2+.+1/2n小于1.
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已知n属于N,n大于等于2,证明:1/2小于1/n+1+1/n+2+.+1/2n小于1.
已知n属于N,n大于等于2,证明:1/2小于1/n+1+1/n+2+.+1/2n小于1.
已知n属于N,n大于等于2,证明:1/2小于1/n+1+1/n+2+.+1/2n小于1.
直接放缩就可以了
1/(n+1)+1/(n+2)+...1/2n
>1/2n+1/2n+.1/2n=1/2
且
1/(n+1)+1/(n+2)+...1/2n
1/(n+1)+1/(n+2)+...+1/2n>1/2n+1/2n+....+1/2n=1/2
1/2<1/(n+1)+1/(n+2)+...+1/2n
①倒序相加法:
设正序和S=1/(n+1)+1/(n+2)+……+1/(2n)
倒序和S'=1/(2n)+1/(2n-1)+……+1/(n+1)
对应相加:
S+S'
=(3n+1)/[(...
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1/(n+1)+1/(n+2)+...+1/2n>1/2n+1/2n+....+1/2n=1/2
1/2<1/(n+1)+1/(n+2)+...+1/2n
①倒序相加法:
设正序和S=1/(n+1)+1/(n+2)+……+1/(2n)
倒序和S'=1/(2n)+1/(2n-1)+……+1/(n+1)
对应相加:
S+S'
=(3n+1)/[(2n)(n+1)]+(3n+1)/[(2n-1)(n+2)]+……+(3n+1)/[(n+1)(2n)]
注意以上n项的通项:(3n+1)/[(2n-i)(n+i+1)]【0<=i<=n-1】
分母部分:
(2n-i)(n+1+i)
=(2n)(n+1)+(2n)i-(n+1+i)i
=(2n)(n+1)+[(n-1)-i]i
>=(2n)(n+1)
∴(3n+1)/[(2n-i)(n+i+1)]<=(3n+1)/[(2n)(n+1)]【i=0或n-1取等号】
∴2S=S+S'
∴S<3/4
②柯西不等式:
S^2=[1/(n+1)+1/(n+2)+……+1/(2n)]^2
<(1^2+1^2+……+1^2)[1/(n+1)^2+1/(n+2)^2+……+1/(2n)^2]【不可能取等号】
=n[1/(n+1)^2+1/(n+2)^2+……+1/(2n)^2]【适当缩小分母部分】
=n[1/n-1/(2n)]
=1/2
∴S<√2/2=2√2/4=√8/4<√9/4=3/4
【高中阶段这题用数学归纳法做最简单。高等数学里当n→∞这是一个有名的级数,和为ln2。而且T=1-1/2+1/3-1/4+……+1/(2n-1)-1/(2n)和题中的S恒等】
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