# What's the most elegant way to show $x_{n+1}=\frac{2x_n^3+a}{3x_n^2}$ converges against $\sqrt[3]{a}$?

Let $1<a\in\mathbb{R}$, $x_0>\sqrt[3]{a}$ and $$\displaystyle x_{n+1}=\frac{2x_n^3+a}{3x_n^2}\;\;\;\;\;(n\in\mathbb{N}_0)$$ It's easy to show that it holds:

1. $x_n>\sqrt[3]{a}$
2. $x_{n+1}<x_n$

Thus, the sequence $(x_n)_n$ converges, since it's monotonically decreasing and lower bounded. However, I want to show that it converges against $\sqrt[3]{a}$. What's the most elegant way to do that?

• I would say that the answer given is the most elegant one given that you already shown that is bounded below and strictly decreasing. Another approach would be showing that the lower bound you found is the greatest lower bound. – Test123 May 7 '14 at 17:41

The following is not elegant, but it is quick. Let the limit be $b$. Then $$b=\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty} \frac{2x_n^3+a}{3x_n^2}=\frac{2b^3+a}{3b^2}.$$ Solve for $b$.
Remark: Let us get more information. Let $b^3=a$, and calculate $x_{n+1}-b$. We have $$x_{n+1}-b=\frac{2x_n^3+b^3}{3x_n^2}-b=\frac{(x_n-b)^2(2x_n+b)}{3x_n^2}.$$ For simplicity imagine that $x_n$ is fairly close to $b$, say $x_n=b\lt 0.1$. The term $\frac{2x_n+b}{3x_n^2}$ is $\lt 1$, so $x_{n+1}-b\lt (0.1)^2$. Do it again. We get $x_{n+2}-b\lt (0.1)^4$, and then $x_{n+3}-b\lt (0.1)^8$, and so on. So once $x_n$ is fairly close to $b$, we get much closer to $b$ dramatically fast. This illustrates the efficiency of the Newton Method.
• Very nice approach! One could phrase it as finding fixed point(s) for the function $$f(x)=\frac{2x^3+a}{3x^2}$$ thus solving $f(b)=b$. To see how $f^n(x)$ converges, one may consider the difference $g(x)=f(x)-x$ which has derivative $$g'(x)=-\frac{2a}{3x^3}-\frac{1}{3}$$ showing that $g(x)$ decreases for positive values of $a$ and $x$. Since $f(b)=b$ implies $g(b)=0$, that $g$ is decreasing means that for $x>b$ we have $g(x)<0$ implying $f(x)<x$. Hence the sequence $f^n(x)$ is decreasing towards the fixed point $b$. – String May 7 '14 at 19:36