Question

Chứng minh rằng:
1/6 < 1/5² + 1/6² + ... + 1/100² < 1/4

Answer 1

\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)

Nên B<\(\dfrac{1}{4}\)

B=\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\)

Nên B>\(\dfrac{1}{6}\)