Cho A=1/3+2/3^2+3/3^3+...+100/3^100 Chứng tỏ rằng: A<¾

=>3A=1+2/3+3/3^2+4/3^3+...+100/3^99 =>3A-A=1+(2/3-1/3)+(3/3^2-2/3^2)+(4/3^3-3/3^3)+...+(100/3^99-99/3^99)-100/3^100 2A=1+(1/3+1/3^2+1/3^3+...+1/3^99)-100/3^100 Đặt B=1/3+1/3^2+1/3^3+...+1/3^99 3B=1+1/3+1/3^2+...+1/3^98 2B=1-1/3^98=>(1-1/3^98):2 Thay vào 2A=>2A=1+1/2-1/3^98:2-100/3^100<1+1/2=3/2 =>A<3/4 Vậy A<3/4