Question

Chứng minh \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2005^2}< \dfrac{2004}{2005}\)

Answer 1

\(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2005^2}\)

\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{20055}\)

\(A< 1-\frac{1}{2005}=\frac{2004}{2005}\)

\(\Rightarrow A< \frac{2004}{2005}\left(đpcm\right)\)

Answer 2

Đặt M=1/2^2+1/3^2+1/4^2+...+1/2005^2

M<1/1.2+1/2.3+1/3.4+...+1/2004.2005

M<1-1/2+1/2-1/3+1/3-1/4+...+1/2004-1/2005

M<1-1/2005=2004/2005(đpcm)