b) Ta có: \(A=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{n\left(n+3\right)}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\)
\(=1-\dfrac{1}{n+3}=\dfrac{n+3-1}{n+3}=\dfrac{n+2}{n+3}< 1\)
