Ta có : \(S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{n\left(n+3\right)}\)
\(\Leftrightarrow S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(S=\frac{1}{1}-\frac{1}{n+3}\)
\(S=\frac{n+3}{n+3}-\frac{1}{n+3}=\frac{n+3-1}{n+3}=\frac{n+2}{n+3}<1\)