An=1/(n+1)+1/(n+2)+.+1/2n,则An+1-An等于?An+1中n+1为下标

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An=1/(n+1)+1/(n+2)+.+1/2n,则An+1-An等于?An+1中n+1为下标
An=1/(n+1)+1/(n+2)+.+1/2n,则An+1-An等于?
An+1中n+1为下标

An=1/(n+1)+1/(n+2)+.+1/2n,则An+1-An等于?An+1中n+1为下标
An=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n)

An+1=1/(n+2)+1/(n+3)+…+1/(2n-1)+1/(2n)+ 1/(2n+1)+1/(2n+2)

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

An=1/(n+1)+1/(n+2) + .... + 1/2n
An+1= 1/(n+2)+1/(n+3)+....+1/2n + 1/(2n + 1) + 1/[2 (n + 1)]
所以 An+1-An = 1/(2n + 1) + 1/[2 (n + 1)] - 1/(n+1) .
= 1 / (2 n + 1) - 1 / [2 (n + 1)]
= 1 / [2 (n + 1) (2 n + 1)]

1/(2n+1)(2n+2)

1/2