Questions about the Inductive sequence. $x_1 = 1$ , $x_{n+1}= \frac{1}{5}(x_n^2+6)$ for $n \geq 1$

Question

I'm a university student and I've got a test coming up and revising. I've answered my question as best I could but if anyone has any tips to set me on the right course that would be fantastic!

The sequence in question is...

$$ x_1 = 1 , x_{n+1}= \frac{1}{5}(x_n^2+6), n \geq 1 $$

The question asks me to show that $x_n < 2$ for all $n\geq 2$. Which I can do. We establish the base case as $x_1 = 1$ which is clearly less than 2 . We assume $x_n < 2$ and prove $x_{n+1}$. We can say $x_n^2 < 4$ so therefore.

$$x_{n+1} = \frac{x_n^2 +6}{5}< \frac{4+6}{5}=2 $$

The question now tells me to prove $x_n$ is increasing.Which I can also demonstrate.

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

$x_n < 2$ and $x_n < 3$ so $(x_n -2)(x_n -3)$ is increasing.

Because the sequence is increasing we can use the monotone convergence theorem to deduce that $\{x_n\}$ converges, so $x_n \rightarrow x $ and $x_{n+1} \rightarrow x$. And by the uniqueness of limits implies...

$$ x= \frac{1}{5}(x^2+6) \implies \frac{1}{5}(x^2 - 5x +6) = 0 \implies \frac{1}{5}(x-2)(x-3) =0 $$

I can only assume (the limit) $x=2$ but I am not sure how to rigorously phrase it instead of intuitively.

Answer 1

Your first step is not entirely correct, it is not necessary that if $x_n<2$ then $x_n^2<4$. What if $x_n=-3$? So you have to prove that $x_n>0$ for all $n$ which is easy to see. So all $0<x_n<2$ and thus limit is $2$ and not $3$.

Answer 2

We know that the limit exists and $x_n<2$, then we have $x=\lim_{n \to \infty}x_n \le 2$.

Suppose on the contrary that we have $x=\lim_{n \to \infty}x_n>2$.

Then $|x-x_n|=|x-2+2-x_n| \ge x-2$,

Then we have $\lim_{n \to \infty}|x-x_n| \ge x-2 > 0$, which is a contradiction since it is suppose to be equal to $0$.

Remark:

Rather than $(x_n-2)(x_n-3)$ is increasing, I think you wanted to say it is positive and hence $x_n$ increases.