Proving convergence of inductively defined sequence

Question

Define a sequence inductively by: $$x_1=1,\space x_{n+1}=\frac{1}{5}(x^2_{n}+6)$$ for $n\ge1$

I need to show that {$x_n$} is an increasing sequence with $x_n \le 2$ for all $n$.

I'm struggling to make a start on this question. I know that I need to use induction but can't think of how to do it.

Answer 1

Clearly $x_1=1<2$, suppose $x_n<2$ then $$x_{n+1}=\frac{1}{5}(x^2_n+6)<\frac{1}{5}(2^2+6)=\frac{10}{5}=2$$ Clearly by definition all terms are positive. Therefore the sequence is bounded. On the other hand note $$x_{n+1}-x_n=\frac{1}{5}(x^2_n+6-5x_n)=\frac{1}{5}(x_n-3)(x_n-2)>0$$ since for all $n$ we showed $x_n<2$. Hence a monotonic increasing sequence. Then by Bolzano-Weierstrass the sequence has a limit point $x$. Thus $$\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\frac{1}{5}(x^2_n+6)\Rightarrow x=\frac{1}{5}(x^2+6)\Rightarrow x^2-5x+6=0\Rightarrow x\in\{2,3\}$$ Since $x_n<2$ for all $n$ then $x=2$ is the solution.

Answer 2

Let $f(x)=\frac{x^2+6}5$. Then $-2<x<2\implies f(x)<2$, because$$f(x)-2=\frac{x^2-4}5<0$$if $x\in(-2,2)$. So, since the first term of your sequence is in $(0,2)$ and since $x_{n+1}=f(x_n)$, every erm of the sequence is in $(0,2)$.

Furthermore,$$f(x)-x=\frac{(x-2)(x-3)}5>0$$if $x\in(0,2)$. So your sequence is strictly increasing.

So, it converges to some $x$ and$$x=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\frac{{x_n}^2+6}5=\frac{x^2+6}5.$$Therefore, $x=3$ or $x=2$. But since every term of the sequence is smallar than $2$, the limit can only be $2$.

Answer 3

Since $x_1=1$, we have $x_2=\frac{1}{5}(x_1^2+6) \le 2$ (using $x_1^2 \le 4)$. Assuming it is true for $x_n$, we have $x_{n+1} \le \frac{10}{5} \le 2$.

Then $$x_{n+1}-x_n=\frac{x_n^2+6-5x_n}{5}=\frac{(x_n-2)(x_n-3)}{5} \ge 0$$