设an=(1/n+1)+(1/n+2)+(1/n+3)+...+1/2n,则an+1-an等于?

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设an=(1/n+1)+(1/n+2)+(1/n+3)+...+1/2n,则an+1-an等于?
设an=(1/n+1)+(1/n+2)+(1/n+3)+...+1/2n,则an+1-an等于?

设an=(1/n+1)+(1/n+2)+(1/n+3)+...+1/2n,则an+1-an等于?
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)]