Question

Chứng minh rằng với mọi số tự nhiên \(n\ge1\) thì

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{n^2}< 1\)

Answer 1

Đặt :

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{n^2}\)

Ta thấy :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

..........................

\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{\left(n-1\right)n}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Leftrightarrow A< 1-\dfrac{1}{n}< 1\)

\(\Leftrightarrow A< 1\)

Vậy ......

Answer 2

A<1 nha ban