Question

A = 1/2² + 1/3² + 1/4² + 1/5² + ... + 1/8² + 1/9² Chứng minh rằng: 2/5 < A < 1

Answer 1

Ta có:

Do \(2^2>1.2\) ; \(3^2>2.3\) ;...; \(9^2>8.9\)

\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

\(\Rightarrow A< 1-\dfrac{1}{9}< 1\) (1)

Lại có: \(2^2< 2.3\) ; \(3^2< 3.4\) ;...; \(9^2< 9.10\)

\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{2}{5}\) (2)

(1);(2) \(\Rightarrow\dfrac{2}{5}< A< 1\)