a,
$u_{n+1}=\dfrac{n+1}{(n+1)^2+1}=\dfrac{n+1}{n^2+2n+2}$
$\Rightarrow u_{n+1}-u_n=\dfrac{n+1}{n^2+2n+2}-\dfrac{n}{n^2+1}=\dfrac{(n+1)(n^2+1)-n(n^2+2n+2)}{(n^2+1)(n^2+2n+2)}=\dfrac{ n^3+n^2+n+1-n^3-2n^2-2n}{(n^2+1)(n^2+2n+2)}=\dfrac{-n^2 - n+1}{(n^2+1)(n^2+2n+2)}=\dfrac{-n^2-n-\dfrac{1}{4} +\dfrac{5}{4} }{(n^2+1)(n^2+2n+2)}=\dfrac{-(n+\dfrac{1}{2})^2+\dfrac{5}{4} }{(n^2+1)(n^2+2n+2)}$
Ta có $-(n+\dfrac{1}{2})^2+\dfrac{5}{4}\le \dfrac{5}{4}$
Dấu bằng xảy ra khi $n=0,75$ (vô lí)
Kể từ $n=1$, $-(1+\dfrac{1}{2})^2+\dfrac{5}{4}=-1<0$
Nên khi $n$ càng tăng, $-(1+\dfrac{1}{2})^2+\dfrac{5}{4}$ càng giảm.
Vậy $-(n+\dfrac{1}{2})^2+\dfrac{5}{4}<0$
$\to u_{n+1}<u_n$
Vậy $(u_n)$ là dãy giảm.
Dễ thấy $u_n\ge 0$ nên $(u_n)$ bị chặn dưới.
$\dfrac{1}{u_n}=n+\dfrac{1}{n}\ge 2\sqrt{n.\dfrac{1}{n}}=2$ (Theo BĐT AM-GM)
$\Rightarrow u_n\le \dfrac{1}{2}$
Vậy $(u_n)$ là dãy bị chặn.
b,
$u_{n+1}=\dfrac{2n+1}{3}$
$\Rightarrow u_{n+1}-u_{n}=\dfrac{2n+1-2n+1}{3}=\dfrac{2}{3}>0$
$\Rightarrow u_{n+1}>u_n$
Vậy $(u_n)$ là dãy tăng.
$n\ge 1\Leftrightarrow 2n-1\ge 1$
$\Rightarrow (u_n)$ bị chặn dưới.
$\lim u_n=\lim\dfrac{2n-1}{3}=\lim\dfrac{2-\dfrac{1}{n}}{\dfrac{3}{n}}=+\infty$
$\Rightarrow (u_n)$ không bị chặn trên.
