Đáp án:
$\begin{array}{l}
{u_n} = \frac{{n - 3}}{{n + 1}}\\
+ {u_1} = \frac{{1 - 3}}{{1 + 1}} = - 1\\
+ {u_2} = \frac{{2 - 3}}{{2 + 1}} = \frac{{ - 1}}{3}\\
+ {u_3} = \frac{{3 - 3}}{{3 + 1}} = 0\\
+ {u_4} = \frac{{4 - 3}}{{4 + 1}} = \frac{1}{5}\\
+ {u_5} = \frac{{5 - 3}}{{5 + 1}} = \frac{2}{6} = \frac{1}{3}\\
Do:{u_2} - {u_1} = - \frac{1}{3} + 1 = \frac{2}{3} > 0\\
\Rightarrow {u_2} > {u_1}
\end{array}$
=> Dãy số tăng.