Question

Cho A =\(\dfrac{1}{1^2.2^2}+\dfrac{1}{2^2.3^2}+\dfrac{1}{3^2.4^2}+........+\dfrac{1}{2016^2.2017^2}vàB=1-\dfrac{1}{2017^2.2018^2}\)

*Hãy so sánh A và B

Answer 1

A < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2016.2017}\)

=> A<\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2016}-\dfrac{1}{2017}\)

=> A<\(1-\dfrac{1}{2017}\)

Vì \(\dfrac{1}{2017}>\dfrac{1}{2017^2.2018^2}\) nên \(1-\dfrac{1}{2017}< 1-\dfrac{1}{2017^2.2018^2}\)

=> A<\(\dfrac{1}{2017}\)<B

Vậy A < B

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