# Thread: sequence

1. ## sequence

Hi everybody

$f(x)=Arctan\frac{1}{x}; x \in [ \frac{\sqrt{3}}{3};1]$
$\{{ U_0=\frac{\sqrt{3}}{3} \atop U_{n+1}=f(U_n)}, \forall n \in \mathbb {N}$

I must show that $(\forall n \in \mathbb{N}): \frac{\sqrt{3}}{3} \le U_n \le 1$ (i used induction but i don't know how to show that: $\frac{\sqrt{3}}{3} \le U_{n+1} \le 1)$

Can you help me please?

2. Originally Posted by lehder
Hi everybody

$f(x)=Arctan\frac{1}{x}; x \in [ \frac{\sqrt{3}}{3};1]$
$\{{ U_0=\frac{\sqrt{3}}{3} \atop U_{n+1}=f(U_n)}, \forall n \in \mathbb {N}$

I must show that $(\forall n \in \mathbb{N}): \frac{\sqrt{3}}{3} \le U_n \le 1$ (i used induction but i don't know how to show that: $\frac{\sqrt{3}}{3} \le U_{n+1} \le 1)$

Can you help me please?
No wonder you didn't succeed: it is false, and already from the beginning!:

$U_1=f(U_0)=f\left(\frac{1}{\sqrt{3}}\right)=Arctan \sqrt{3} =\frac{\pi}{3}>1$...

Tonio