Question

let x1=1 and Xn+1=squrrt(3xn)for all n€N show that (xn) is bounded ,monotone and converges to 3

Answer 1

Let $f(x)=\sqrt{3x}$ for all $x>0$, then we have $x_{n+1}=f(x_n)$ for all $n\in\mathbb{N}$.

Since $f(x)$ is a monotonously increasing function, then any recurrent sequence satisfying $x_{n+1}=f(x_n)$ is monotone. Really:

If $x_0\leq x_1$ then $f(x_0)\leq f(x_1)$, that is, $x_1\leq x_2$ and, by induction, $x_n\leq x_{n+1}$.

If $x_0\geq x_1$ then $f(x_0)\geq f(x_1)$, that is, $x_1\geq x_2$ and, by induction, $x_n\geq x_{n+1}$.

By the condition, $x_0=1$, thus, $x_1=\sqrt{3}>1=x_0$, therefore, this sequence is monotonously increasing. Note that if for any $n$, $x_n< 3$ then $x_{n+1}=f(x_n)<f(3)=3$, and, by induction, $x_n< 3$ for all $n\in\mathbb{N}$. Thus, the given sequence is increasing and bounded from above by 3. Therefore, $x_n$ is convergent. Let $a\in\mathbb{R}$ is limit of this sequence.

Since $a=\lim\limits_{n\to+\infty}x_n$ and $f(x)$ is continuous everywhere on $(0,+\infty)$, we have

$f(a)=\lim\limits_{n\to+\infty}f(x_n)=\lim\limits_{n\to+\infty}x_{n+1}=a$

$\sqrt{3a}=a$ implies $a=0$ or $a=3$. Since $x_n$ is increasing, $\lim\limits_{n\to+\infty}x_n\geq x_0=1$, therefore, $a\ne 0$, and hence, $a=3$.