f(n)=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n) (n≥2,n∈N*)
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f(n)=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n) (n≥2,n∈N*)
判断f(n)单调性 求它的最值
判断f(n)单调性 求它的最值
![f(n)=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n) (n≥2,n∈N*)](/uploads/image/z/5509367-71-7.jpg?t=f%28n%29%3D1%2F%28n%2B1%29%2B1%2F%28n%2B2%29%2B%E2%80%A6%2B1%2F%282n-1%29%2B1%2F%282n%29+%EF%BC%88n%E2%89%A52%2Cn%E2%88%88N%2A%EF%BC%89)
f(n)=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n)
则
f(n+1)=1/(n+2)+1/(n+3)+…+1/(2n-1)+1/(2n)+ 1/(2n+1)+1/(2n+2)
则
f(n+1)-f(n)
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)
=1/[(2n+1)(2n+2)]
>0
所以f(n)为单增函数.有最小值.当n=2时取得最小值
f(2)=1/3+1/4=7/12
则
f(n+1)=1/(n+2)+1/(n+3)+…+1/(2n-1)+1/(2n)+ 1/(2n+1)+1/(2n+2)
则
f(n+1)-f(n)
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)
=1/[(2n+1)(2n+2)]
>0
所以f(n)为单增函数.有最小值.当n=2时取得最小值
f(2)=1/3+1/4=7/12
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