Vì \(\dfrac{1}{a}\left(a>1\right)< 1với\forall a\)
mà \(2^2;3^2;.....;100^2>1\)
\(=>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)
