Bài này chứng minh hay tính tổng thật đó ?
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..+\dfrac{1}{99.100}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{1}-\dfrac{1}{100}=\dfrac{99}{100}< 1\) Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
Đặt : \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
....................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{1}{100}\)
Mà \(\dfrac{1}{100}< \dfrac{100}{100}=1\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{100}{100}=1\)(đpcm)
