Giải thích các bước giải:
Đặt `A = 1/3 + 2/3^2 + 3/3^3 + ... + 2019/3^2019`
`=> 3A = 1 + 2/3 + 3/3^2 + ... + 2019/3^2018`
`=> 3A - A = (1 + 2/3 + 3/3^2 + ... + 2019/3^2018) - (1/3 + 2/3^2 + 3/3^3 + ... + 2019/3^2019)`
`=> 2A = 1+ 1/3 + 1/3^2 + ... + 1/3^2018 - 2019/3^2019`
`=> 2A < 1+ 1/3 + 1/3^2 + ... + 1/3^2018 (1)`
Đặt `B = 1+ 1/3 + 1/3^2 + ... + 1/3^2018`
`=> 3B = 3 + 1 + 1/3 + ... + 1/3^2017`
`=> 3B - B = (3 + 1 + 1/3 + ... + 1/3^2017)- (1+ 1/3 + 1/3^2 + ... + 1/3^2018)`
`=> 2B = 3 - 1/3^2018`
`=> B = 3/2 - 1/3^2018 . 2 `
`=> B < 3/2 (2)`
Từ `(1)` và `(2) => 2A < 3/2`
`=> A < 3/4 = 0,75 (đpcm)`
