Ta có :
\(\dfrac{1}{3^2}=\dfrac{1}{3\times3}< \dfrac{1}{2\times3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4\times4}< \dfrac{1}{3\times4}\)
Tương tự như vậy ta cũng được \(\dfrac{1}{100}=\dfrac{1}{10\times10}< \dfrac{1}{9\times10}\)
Do đó :
\(A< \dfrac{1}{2\times3}+\dfrac{1}{3\times4}+.....+\dfrac{1}{9\times10}\)
Suy ra \(A< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}< \dfrac{1}{2}\)
Do đó \(A< \dfrac{1}{2}\left(đpcm\right)\)