Question

A=\(\dfrac{1}{2^2}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{4^2}\)+ .....+\(\dfrac{1}{2008^2}\). Chứng tỏ rằng: A<1

Answer 1

Đặt \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)

Ta có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2008^2}< \)\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2007\cdot2008}\left(1\right)\)

Lại có: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)

\(=1-\dfrac{1}{2008}< 1\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) ta có \(A< B< 1\Rightarrow A< 1\)

Answer 2

A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{2008^2}\)

A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2007.2008}\)

A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)

A<\(1-\dfrac{1}{2008}\)

A<\(\dfrac{2007}{2008}< 1\)

=> A<1

Vậy A<1