Help with convergence of a recursive sequence

Question

We define a sequence of real numbers $\{ x_{n}\} $ in a recursive form where $ x_{0} = 0$ and $$ x _{n} = \frac{x_{n-1} ^2+2}{3}$$ for $ n \geq 1.$

We need to prove that this sequence converges and to calculate $\lim\limits_{n\to \infty} x_{n}.$

I think I know how to do the second part, the part of the limit.
Suppose that $\lim\limits_{n\to \infty} x_{n} = L .$ Then $ L = \frac{L^2 +2}{3}$ and we find the solution for $L$ (the possible solutions are $L =1$ or $L =2$).

But of course, all of this depends on the first part, showing the convergence of the sequence, and I don't know how to begin with this particular sequence.
All I can see is that $ x_{0} \lt x_{1} ,$ that $\frac {2}{3}$ is a lower bound of the sequence and that the sequence slowly increases (evidently, at some point the sequence increases so little that it has an upper bound at $L = 1$ or $2,$ most probably $2$).

I would appreciate some suggestions on how to solve it. Thanks!

Answer 1

As a hint, try to show that

• $0 \leq x_n < 1$ for any $n$;
• $x_{n + 1} > x_n$ for any $n$.

Prove the above statements by induction on $n$.

Answer 2

A simple inductive argument shows $x_n<1$ for all $n$. Further, note that $$x_{n+1}-x_n=\frac{x_n^2+2}{3}-x_n=\frac{(x_n-1)(x_n-2)}{3}>0$$ Thus, $\{x_i\}$ is a bounded increasing sequence and hence, converges.

Answer 3

It is clear that $\forall n\in \mathbb{N}, x_n>0.$

Let us prove that the sequence is increasing.
For any $n\geq 2$ is $$x_{n+1}-x_n=\frac{(x_{n}-x_{n-1})(x_{n}+x_{n-1})}{3}$$ Hence the difference $(x_{n+1}-x_n)$ has constant sign, which is positive because $x_1>x_0.$ The sequence is increasing.

Now, $x_0<1.$ By induction, from $x_n<1$ we deduce $$x_{n+1}<\frac{1+2}{3}=1.$$

The sequence is increasing and bounded from above by $1.$