Question

CHo \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2020^2}\) chứng tỏ A<1

Answer 1

`1/2^2 < 1/(1.2)`

`1/3^3 < 1/(2.3)`

`...`

`1/(2020^2) < 1/(2019.2020)`

`=> A < 1/(1.2) + 1/(2.3) + ... + 1/(2019.2020)`

`=> A < 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/2019 - 1/2020 < 1`.

Answer 2

\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2020^2}< \dfrac{1}{2019.2020}\)

Vậy \(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2019.2020}=\dfrac{1}{1}-\dfrac{1}{2020}=\dfrac{2019}{2020}< \dfrac{2020}{2020}=1\)