Question

Cho tổng A = \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+...+ \(\frac{1}{2011^2}\). Chứng tỏ A < 1

Answer 1

Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\)

Ta có:

\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\left(1\right)\)

Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}< 1\left(2\right)\)

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

Đpcm