# The lower bound for the following sequence.

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?

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 $$n$$).
• Can not we say that it is bounded below by $2/3$? Oct 14, 2022 at 18:47
• Yes, the sequence is also bounded below by $\frac 23$, because I have shown that it is even bounded by $1$. But $\frac 23$ is not the best possible bound Oct 23, 2022 at 5:44