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

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

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

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

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

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