Question

Hãy so sánh 1/2² + 1/3² + 1/4² + ....+1/2025² với 1 .giúp e với ạ

Answer 1

Đặt A = 1/ 2^2 + 1/3^2 + 1/4^2 + ... + 1/2025^2

A < 1/1 * 2 + 1/2 * 3 + 1/3 * 4 + ... + 1/2024 * 2025

A < 1/1 - 1/2 + 1/2- 1/3 + 1/3 - 1/4 + ...+ 1/2024 - 1/2025

A < 1- 1 / 202

=> A < 1

Answer 2

Mk nhầm

Answer 3

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\)

Answer 4

Đặt `A = 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(2025^2)`

Ta có : `1/(2^2) <1/(1*2)`

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

`...........`

`1/(2025^2) < 1/(2024 *2025)`

`=> 1/(2^2) +1/(3^2) + ... + 1/(2025^2) < 1/(1*2) + 1/(2*3)+ ... + 1/(2024*2025)`

`=> A < 1/(1*2) + 1/(2*3)+ ... + 1/(2024*2025)`

`=> A< 1/1 - 1/2 + 1/2 -1/3 + ... +1/2024 - 1/2025`

`=> A < 1 - 1/2025`

`=> A <1 ` (VÌ `1 - 1/2025 <1`)

Vậy `A<1`