lim(1+1/3+1/9+.+1/3^n)=

lim(1+1/3+1/9+.+1/3^n)=
lim(1+1/3+1/9+.+1/3^n)=

lim(1+1/3+1/9+.+1/3^n)=
设Sn=1+1/3+1/3²+.+1/3^n.(1)
∵1/3Sn=1/3+1/3²+1/3³+.+1/3^n+1/3^(n+1).(2)
∴由(1)式-(2)式得(1-1/3)Sn=1-1/3^(n+1)
==>Sn=[1-1/3^(n+1)]/(1-1/3)
=(3/2)[1-1/3^(n+1)]
故lim(n->∞)(1+1/3+1/3²+.+1/3^n)=lim(n->∞)Sn
=lim(n->∞){(3/2)[1-1/3^(n+1)]}
=(3/2)(1-0)
=3/2.