公比q=10,首项a1=3
an=3*10^(n-1)
则bn=1/{lg[3*10^(n-1)]*lg[3*10^n]}
=1/[(n-1+lg3)(n+lg3)]
=1/(n-1+lg3)-1/(n+lg3)
所以{bn}的各项的和Sn=[1/lg3-1/(1+lg3)]+[1/(1+lg3)-1/(2+lg3)]+.+[1/(n-1+lg3)-1/(n+lg3)]
=1/lg3-1/(n+lg3)
=n/[lg3(n+lg3)]