若Sn=n乘(n+2),怎么才能证明1/S1+2/S2+3/S3+…+1/Sn<3/4?

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解决时间 2021-02-20 10:12

提问者网友：两耳就是菩提
2021-02-19 13:05

最佳答案

五星知识达人网友：老鼠爱大米
2021-02-19 14:01

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

1/S1+1/S2+…+1/Sn
=[(1-1/3)+(1/2-1/4)+(1/3-1/5)+...+1/(n-1)-1/(n+1)+1/n-1/(n+2)]/2
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-[1/(n+1)+1/(n+2)]/2
所以1/S1+1/S2+…+1/Sn<3/4

全部回答

1楼网友：鱼忧
2021-02-19 16:15

Sn=n*(n+2) ,则1/Sn=1/2[1/n-1/(n+2)] 所以 1/S1+2/S2+3/S3+…+1/Sn =1/2[1/1-1/3+1/2-1/4+1/3-1/5...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)] 化简得3/4-1/2[1/(n+1)+1/(n+2)]＜3/4

2楼网友：青尢
2021-02-19 15:16

Sn=n(n+2) 1/Sn=1/[n(n+2)]=(1/2)[1/n-1/(n+2)] 1/S1+1/S2+...Sn=(1/2)[1-1/3+1/2-1/4+1/3-1/5+…+1/(n-1)-1/(n+1)+1/n-1/(n+2)]=(1/2)[1+1/2-1/(n+1)-1/(n+2)]=3/4-(2n+3)/[2(n+1)*(n+2)] 很明显是小于3/4的

3楼网友：荒野風
2021-02-19 14:06

sn=(3+2n+1)n/2=n(n+2)

tn=1/s1+2/s2+3/s3+……+n/sn

=1/(1*3)+1/(2*4)+1/(3*5)+...+1/[(n-1)(n+1)+1/[n(n+2)]

=1/2×[(1-1/3)+(1/2-1/4)+(1/3-1/5)+..+[1/(n-1)-1/(n+1)]+[1/n-1/(n+2)]

=1/2×[1+1/2-1/(n+1)-1/(n+2)]

=3/4-(2n+3)/[2(n+1)(n+2)]

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