Question

cmr : 1/2^2+1/3^2+.....+1/2017^2< 2

Answer 1

Giải:

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2017^2}\)

Dễ thấy:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(....\)

\(\dfrac{1}{2017^2}=\dfrac{1}{2017.2017}< \dfrac{1}{2016.2017}\)

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

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

\(\Rightarrow A< 1-\dfrac{1}{2017}< 2\)

\(\Rightarrow A< 2\left(dpcm\right)\)