Bài 4:

`a)` Ta có:

`A = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2 < 1`

`1/2^2 < 1/(1*2)` `1/3^2 < 1/(2*3)`

`1/4^2 < 1/(3*4)` `...` `1/100^2 < 1/(99*100)`

`A = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2 < 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100)`

`A < (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/99 - 1/100)`

`A < 1 - 1/100`

`A < 99/100`

Vì `99/100 < 1` nên `A < 1` `("đpcm")`

`b)` Ta có:

`B = 1/3 + 2/3^2 + 3/3^3 + 4/3^4 + ... + 100/3^100 + 101/3^101 < 3/4`

`B = 1/3 + 2/3^2 + 3/3^3 + ... + 101/3^101`

`3B = 3 * (1/3 + 2/3^2 + 3/3^3 + ... + 101/3^101)`

`3B = 1 + 2/3 + 3/3^2 + ... + 101/3^100`

`3B- B= (1 + 2/3 + 3/3^2 + ... + 101/3^100) - (1/3 + 2/3^2 + 3/3^3 + ... + 101/3^101)`

`2B = 1 + (2/3 - 1/3) + (3/3^2 - 2/3^2) + ... + (101/3^100 - 100/3^100) - 101/3^101`

`2B = 1 + 1/3 + 1/3^2 + ... + 1/3^100 - 101/3^101`

Đặt `C = 1 + 1/3 + 1/3^2 + ... + 1/3^100`

`3C = 3 * (1 + 1/3 + 1/3^2 + ... + 1/3^100)`

`3C = 3 + 1 + 1/3 + ... + 1/3^99`

`3C-C= (3 + 1 + 1/3 + ... + 1/3^99) - (1 + 1/3 + 1/3^2 + ... + 1/3^100)`

`2C = 3 - 1/3^100`

`C = (3 - 1/3^100)/2 = 3/2 - 1/(2 * 3^100)`

Thay `C` vào `2B` ta được:

`2B = C - 101/3^101`

`2B = (3/2 - 1/(2 * 3^100)) - 101/3^101`

`2B = 3/2 - (1/(2 * 3^100) + 101/3^101)`

Vì `1/(2 * 3^100) > 0` và `101/3^101 > 0` nên `(1/(2 * 3^100) + 101/3^101) > 0`

Do đó, `2B < 3/2` hay `B < 3/2 : 2 <=>B < 3/4` `("đpcm")`