数列an的前n项的和Sn=(1/3)*an-2,求:lim(a1+a2+...+an)

S1=a1=1/3*a1-2 a1= - 3
a(n)=S(n)-S(n-1)=(1/3)*[a(n)-a(n-1)]
(2/3)*a(n)=-1/3*a(n-1)
a(n)=-1/2*a(n-1)
等比数列,公比为-1/2
a(n)=(-1/2)^(n-1)*(-3)
a1+a2+...an=sn=1/3*a(n)-2=(-1/2)^(n-1)*(-1)-2
lim(a1+a2+...+an)=lim(-1/2)^(n-1)*(-1)-2=-2