Question

\(choA=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{n^2}+...+\dfrac{1}{2004^2}\)

\(chứngminh\dfrac{1}{65}< A< \dfrac{1}{4}\)

Answer 1

\(A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{2003.2004}\)

\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\)

\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{2004}< \dfrac{1}{4}\)

Đồng thời:

\(A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2004.2005}\)

\(A>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2004}-\dfrac{1}{2005}\)

\(A>\dfrac{1}{5}-\dfrac{1}{2005}=\dfrac{80}{401}>\dfrac{50}{500}>\dfrac{1}{10}>\dfrac{1}{65}\)

Vậy \(\dfrac{1}{65}< A< \dfrac{1}{4}\)