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(1/2)谢谢解答下列问题祝中秋快乐.设f(n)=(1/n+1)+(1/n+2)+(1/n+3)+…+(1/2n)(n
人气:233 ℃ 时间:2020-09-13 13:43:05
解答
f(n)有n项,则
f(n+1)-f(n)
=[(1/(n+2))+(1/(n+3))+···+(1/(2n))+(1/(2n+1))+(1/(2n+2))]
-[(1/(n+1))+(1/(n+2))+···+(1/(2n))]
=[(1/(2n+1))+(1/(2n+2))]-(1/(n+1))
=(1/(2n+1))-(1/(2n+2))
=1/(2(n+1)(2n+1))
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