Question

Xét tính tăng giảm của dãy số: u_n = \(\sqrt{n+10}-\sqrt{n+2}\)

Answer 1

\(u_n=\sqrt[]{n+10}-\sqrt[]{n+2}\)

\(\Leftrightarrow u_n=\dfrac{n+10-\left(n+2\right)}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(\Leftrightarrow u_n=\dfrac{8}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(u_{n+1}=\sqrt[]{n+11}-\sqrt[]{n+3}\)

\(\Leftrightarrow u_{n+1}=\dfrac{n+11-\left(n+3\right)}{\sqrt[]{n+11}+\sqrt[]{n+3}}\)

\(\Leftrightarrow u_{n+1}=\dfrac{8}{\sqrt[]{n+11}+\sqrt[]{n+3}}\)

\(u_{n+1}-u_n=8\left(\dfrac{1}{\sqrt[]{n+11}+\sqrt[]{n+3}}-\dfrac{1}{\sqrt[]{n+10}+\sqrt[]{n+2}}\right)\)

mà \(\dfrac{1}{\sqrt[]{n+11}+\sqrt[]{n+3}}< \dfrac{1}{\sqrt[]{n+10}+\sqrt[]{n+2}}\)

\(\Rightarrow u_{n+1}-u_n< 0\)

Vậy dãy đã cho là dãy số giảm