Question

Chứng minh rằng :

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}+\dfrac{1}{2022^2}< 1\)

Answer 1

Ta thấy: \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}\)

\(...\)

\(\dfrac{1}{2021^2}< \dfrac{1}{2020\cdot2021}\)

\(\dfrac{1}{2022^2}< \dfrac{1}{2021\cdot2022}\)

Suy ra: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dots+\dfrac{1}{2021^2}+\dfrac{1}{2022^2}\)

\(< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dots+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}< 1\left(dpcm\right)\)

Answer 2

tui gõ word

Answer 3