因为an=2n+1
所以{an}是等差数列
所以Sn=n(a1+an)/2=n(3+2n+1)/2=n(n+2)
所以1/Sn=1/n(n+2)=[1/n-1/(n+2)]/2
所以数列{1/Sn}的前n项和Tn=S1+S2+...+Sn
=[1/1-1/(1+2)]/2+[1/2-1/(2+2)]/2+...+[1/n-1/(n+2)]/2
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-(2n+3)/[2(n+1)(n+2)]