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
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