Question

Xét tính tăng giảm của dãy \(\left(u_n\right):\left\{{}\begin{matrix}u_1=\dfrac{1}{\sqrt{3}}\\u_{n+1}=\dfrac{\sqrt{u_n^2+1}-1}{u_n}\left(n=1;2;3;...\right)\end{matrix}\right.\)

Answer 1

Với n = 1 => \(u_2=\dfrac{\sqrt{\dfrac{1}{3}+1}-1}{\dfrac{1}{\sqrt{3}}}=\dfrac{\dfrac{2}{\sqrt{3}}-1}{\dfrac{1}{\sqrt{3}}}=\dfrac{\dfrac{2-\sqrt{3}}{\sqrt{3}}}{\dfrac{1}{\sqrt{3}}}=2-\sqrt{3}\)

Ta có \(u_2-u_1< 0\Rightarrow u_2< u_1\)

=> (u_n) giảm dần