c)
$\begin{array}{l} {u_n} = \dfrac{{2n}}{{{n^2} + 1}}\\ {u_{n + 1}} = \dfrac{{2\left( {n + 1} \right)}}{{{{\left( {n + 1} \right)}^2} + 1}}\\ \Rightarrow {u_{n + 1}} - {u_n} = \dfrac{{2\left( {n + 1} \right)}}{{{n^2} + 2n + 2}} - \dfrac{{2n}}{{{n^2} + 1}}\\ = \dfrac{{2\left( {n + 1} \right)\left( {{n^2} + 1} \right) - 2n\left( {{n^2} + 2n + 2} \right)}}{{\left( {{n^2} + 2n + 2} \right)\left( {{n^2} + 1} \right)}}\\ = \dfrac{{2\left( {{n^3} + n + {n^2} + 1} \right) - 2{n^3} - 4{n^2} - 4n}}{{\left( {{n^2} + 2n + 2} \right)\left( {{n^2} + 1} \right)}}\\ = \dfrac{{ - 2{n^2} - 2n + 2}}{{\left( {{n^2} + 2n + 2} \right)\left( {{n^2} + 1} \right)}} = - \dfrac{{2\left( {{n^2} + n - 1} \right)}}{{\left( {{n^2} + 2n + 2} \right)\left( {{n^2} + 1} \right)}}\\ {n^2} + n - 1 > 0 \Leftrightarrow \left[ \begin{array}{l} n < - \dfrac{{1 + \sqrt 5 }}{2}\\ \dfrac{{ - 1 + \sqrt 5 }}{2} < n \end{array} \right.\\ n \in \mathbb{N*}\Rightarrow n \ge 1 > \dfrac{{ - 1 + \sqrt 5 }}{2} \Rightarrow {n^2} + n - 1 > 0\\ \Rightarrow - \dfrac{{2\left( {{n^2} + n - 1} \right)}}{{\left( {{n^2} + 2n + 2} \right)\left( {{n^2} + 1} \right)}} < 0\\ \Rightarrow {u_{n + 1}} - {u_n} < 0 \Rightarrow {u_{n + 1}} < {u_n} \end{array}$
Vậy đây là dãy số giảm.
$\begin{array}{l}
n \in \mathbb{N*} \Rightarrow n > 0 \Rightarrow {u_n} = \dfrac{{2n}}{{{n^2} + 1}} > 0\\
{n^2} + 1 \ge 2n \Rightarrow \dfrac{{2n}}{{{n^2} + 1}} \le \dfrac{{2n}}{{2n}} = 1\\
\Rightarrow 0 < {u_n} \le 1
\end{array}$
Vậy đây là dãy số bị chặn
