设等比数列{an}的公比为q
侧：S n=a1(q的n次方-1)/(q-1)
Tn=1/a1+1/a2+,=1/a1[((1/q)的n次方-1)/(1/q-1)
= [(q的n次方-1)/(q-1)]/[ a1• q的(n-1`)次方]
=[(q的n次方-1)/(q-1)]/(an)
故：Sn/Tn=a1•an