已知数列{An}的前n项和Sn满足Sn=1-2/3An.

n=1时 S1=1-(2/3)A1 解得A1=3/5
n>1时 S(n-1)=1-(2/3)A(n-1)
An=Sn-S(n-1)=-(2/3)An+(2/3)A(n-1)
所以An=(2/5)*A(n-1)
所以{An}是公比为2/5的等比数列
故An=(3/5)*(2/5)^(n-1)
所以Sn=1-(2/3)An
=1-(2/5)*(2/5)^(n-1)
=1-(2/5)^n
AnSn=An*[1-(2/3)An]
=An-(2/3)An²
=An-(2/3)*(3/5)²*(2/5)^(2n-2)
=An-(6/25)*(4/25)^(n-1)
所以A1S1+A2S2+...+AnSn
=(A1+A2+.+An)-(6/25)*[1+(4/25)+...+(4/25)^(n-1)]
=Sn-(6/25)*[1-(4/25)^n]/(1-4/25)
=1-(2/5)^n-2/7+(2/7)*(4/25)^n
=5/7-(2/5)^n-(4/25)^n
故lim(A1S1+A2S2+…AnSn) (n→∞)
=5/7