问题:

【若函数f(n)=1/(n+1)+1/(n+2)+...+1/(n+n),n>=2,n为正整数,求函数的最小值n-[n/(n+1)+(n+1)/(n+2)+....+(n+n-1)/(n+n)]≥n-n·{[n/(n+1)][(n+1)/(n+2)]....[(n+n-1)/(n+n)]}^(1/n)这一步是怎么来的啊?看不懂.】

更新时间:2024-04-19 08:32:25

问题描述:

若函数f(n)=1/(n+1)+1/(n+2)+...+1/(n+n),n>=2,n为正整数,求函数的最小值

n-[n/(n+1)+(n+1)/(n+2)+....+(n+n-1)/(n+n)]

≥n-n·{[n/(n+1)][(n+1)/(n+2)]....[(n+n-1)/(n+n)]}^(1/n)

这一步是怎么来的啊?看不懂

.

雷小峰回答:

  "夹逼定理"

  f(n)=1/(n+1)+1/(n+2)+...+1/(n+n)

  =[1-n/(n+1)]+[1-(n+1)/(n+2)]+...+[1-(n+n-1)/(n+n)]

  =n-[n/(n+1)+(n+1)/(n+2)+.+(n+n-1)/(n+n)]

  ≥n-n·{[n/(n+1)][(n+1)/(n+2)].[(n+n-1)/(n+n)]}^(1/n)

  =n-n·(1/2)^(1/n)

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