a) $U_{n+1}-U_{n} = \dfrac{2 - (n+1)}{\sqrt{n+1}}-\dfrac{2 - n}{\sqrt{n}}= \left ( \dfrac{2}{\sqrt{n+1}}-\sqrt{n+1}\right )- \left ( \dfrac{2}{\sqrt{n}}-\sqrt{n}\right )$$=2\left ( \dfrac{1}{\sqrt{n+1}}-\dfrac{1}{\sqrt{n}} \right )+\left ( \sqrt{n}-\sqrt{n+1}\right )$ do $\begin{cases} \dfrac{1}{\sqrt{n+1}}<\dfrac{1}{\sqrt{n}} \\\sqrt{n}<\sqrt{n+1} \end{cases}\Rightarrow U_{n+1}-U_{n}<0\Rightarrow U_{n+1}<U_{n}$ Vậy dãy đã cho là dãy giảm.
a) $U_{n+1}-U_{n} = \dfrac{2 - (n+1)}{\sqrt{n+1}}-\dfrac{2 - n}{\sqrt{n}}= 2\left ( \dfrac{1}{\sqrt{n+1}}-\dfrac{1}{\sqrt{n}} \right )+\left ( \sqrt{n}-\sqrt{n+1}\right )$ do $\begin{cases} \dfrac{1}{\sqrt{n+1}}<\dfrac{1}{\sqrt{n}} \\\sqrt{n}<\sqrt{n+1} \end{cases}\Rightarrow U_{n+1}-U_{n}<0\Rightarrow U_{n+1}<U_{n}$ Vậy dãy đã cho là dãy giảm.
a) $U_{n+1}-U_{n} = \dfrac{2 - (n+1)}{\sqrt{n+1}}-\dfrac{2 - n}{\sqrt{n}}=
\left ( \dfrac{2}{\sqrt{n+1}}-\sqrt{n+1}\right )- \left ( \dfrac{2}{\sqrt{n}}-\sqrt{n}\right )$$=2\left ( \dfrac{1}{\sqrt{n+1}}-\dfrac{1}{\sqrt{n}} \right )+\left ( \sqrt{n}-\sqrt{n+1}\right )$ do $\begin{cases} \dfrac{1}{\sqrt{n+1}}<\dfrac{1}{\sqrt{n}} \\\sqrt{n}<\sqrt{n+1} \end{cases}\Rightarrow U_{n+1}-U_{n}<0\Rightarrow U_{n+1}<U_{n}$ Vậy dãy đã cho là dãy giảm.