Question

Chứng minh rằng cới mọi số tự nhiên \(n\ge2\):

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

Answer 1

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

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

\(\Rightarrow2A< \dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(n-1\right)\left(n+1\right)}\)

\(\Rightarrow2A< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n+1}\)

\(\Rightarrow2A< 1\)

\(\Rightarrow A< \dfrac{1}{2}< \dfrac{2}{3}\)