The lower bound for the following sequence.

Question

Assume a sequence $\{x_n\}_{n=1}^{\infty}$ is defined recursively by $x_1 = 2$ and $$x_{n+1} = x_n - \frac{(x_n)^3 - x_n}{3 (x_n)^2 - 1} \quad \quad \forall n \geq 1.$$

Find the limit of $x_n$ as $n \to \infty.$

My trial:

I was able to show that the sequence is decreasing but I was not able to show that it is bounded below so that it will be convergent. Can someone show me how it is bounded below please?

Answer 1

Spoiler: The sequence is bounded below by $1$.

I will provide a detailed proof, if you want to try yourself, maybe don't look at my answer at first: We show that by induction. Let $x_n \ge 1$. Then $(x_n)^3-x_n\ge 0$ and $3(x_n)^2-1 >0$. Then $$x_{n+1} = \frac{3x_n^3-x_n}{3x_n^2-1} - \frac{x_n^3-x_n}{3x_n^2-1} = 2 \frac{x_n^3}{3x_n^2-1}$$ The last expression is a function in $x_n$ with the following properties: $f(x) = 2 \frac{x^3}{3x^2-1}$ has the derivative $$f'(x) = \frac{6x^2 (3x^2-1) - 6x \cdot 2x^3}{(3x^2-1)^2}=\frac{6x^4-6x^2}{(3x^2-1)^2}$$ which is positive for all $x>1$ because the denominator is non-zero (its zeros are at $\pm \frac{1}{\sqrt 3}$) and the numerator is $6x^2(x^2-1)$ which will be positive for $x>1$ as well. Therefore $f(1) < f(x)$ for all $x>1$. As $f(1) = 1$ we conclude that $x_{n+1} = f(x_n) \ge 1$.

For finding the limit we realise that $x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}$ for the function $g(x) = x^3-x$. Because the derivative satisfies $$|g'(x)| \ge 2$$ for all $x \ge 1$ the denominator will never vanish. We can conclude that the limit $l$ has to satisfy $$l= l - \frac{g(l)}{(g'(l)}$$ which means that $g(l) = 0$. Therefore $l^3=l$ which is the case for $l=1$ (and $l=0$ or $l=-1$ but these two can't be the limit because $x_n \ge 1$ for all $n$).