所以,
1/an≤[1/3^(n-1)]
(1/a1)+(1/a2)+.......+(1/an)≤1+(1/3)+(1/3^2)+.....+[1/3^(n-1)]=[1/(1-1/3)[1-(1/3)^n]<[1/(1-1/3)][1-0]=3/2
(1/a1)+(1/a2)+.......+(1/an)<3/2