S= $\frac{3}{1.4}$ + $\frac{3}{4.7}$ + $\frac{3}{7.10}$ +...+ $\frac{3}{n + (n+3)}$
= $\frac{1}{1}$ - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{7}$ + $\frac{1}{7}$ - $\frac{1}{10}$ +...+ $\frac{1}{n.(n+3)}$
= $1$ - $\frac{1}{n+3}$
= $\frac{n+3}{n+3}$ - $\frac{1}{n+3}$
= $\frac{n+3-1}{n+3}$ = $\frac{n+2}{n+3}$
Vì n ∈ N* ⇒ n+2 < n+3
⇒ $\frac{n+2}{n+3}$ < 1
⇒ S < 1 (đpcm)