Question

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

Answer 1

ta thấy : \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}=1-\dfrac{1}{2}\)

tương tự: \(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

....

\(\dfrac{1}{2005^2}=\dfrac{1}{2005.2005}< \dfrac{1}{2004.2005}=\dfrac{1}{2004}-\dfrac{1}{2005}\)

cộng vế theo vé các BĐT trên, ta có:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2005^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2004}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)=> đpcm