Question

Chứng minh :

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

 

Answer 1

Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}\)

Ta có:

\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}\left(1\right)\)

Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}\)

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

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

Từ (1) và (2) \(\Rightarrow A< B< 1\Rightarrow A< 1\) (đpcm)