Convergence of the recursive sequence $x_n=\frac{x_{n-1}}2+\frac3{x_{n-1}},n\ge 1$

Question

Consider the iterative scheme :

$x_n=\frac{x_{n-1}}2+\frac3{x_{n-1}},n\ge 1$

with the initial point $x_0 \gt 0$. Then the sequence $\{x_n\}$

$(a)$ converges only if $x_o \lt 3$

$(b)$converges for any $x_0$

$(c)$does not converge for any $x_0$

$(d)$converges only if $x_0\gt 1$

My attempt :

Since $x_0\gt 0$ ,we have $x_n\gt 0, \forall n\in \mathbb{N}$

Now , $x_n-x_{n-1}=\frac{x_{n-1}^2+6-2x_{n-1}^2}{x_{n-1}}=\frac{6-x_{n-1}^2}{2x_{n-1}}$

$\Rightarrow x_n\lt ,\gt x_{n-1}$ according as $x_{n-1}\gt ,\lt \sqrt{6}$.

Again,by the given recurrence relation ,

$x_{n-1}^2-2x_nx_{n-1}+6=0$

This is quadratic in $x_{n-1}$ and to have a real solution

$4x_{n}^2-24\ge 0$ i.e $x_n\ge \sqrt6$ or $x_{n}\le -\sqrt6$

The latter is not possible, so $x_n\gt \sqrt6$

What does this mean?? Does this mean that even if $x_0$ be any value greater than zero and smaller than $\sqrt6$, the sequence is ultimately monotone decreasing converging to $\sqrt6$ (limit obtained from the relation)? So, should my answer be $(b)$?

Please help me with your ideas. Thanks for your time.

Answer 1

If $x_0>0$ we have $x_n>0$ and from here: $$x_n-\sqrt6=\frac{(x_{n-1}-\sqrt6)^2}{2x_{n-1}}\geq0,$$ which by your work: $$x_n-x_{n-1}=\frac{6-x_{n-1}^2}{2x_{n-1}}\leq0$$ gives that $x$ decreases.

Id est, our sequence is closed to $\sqrt6$ for any $x_0>0$.

Answer 2

Since finite terms of the sequence don't matter, so even if $x_0>\sqrt{6},$ the sequence still remains monotonically decreasing and bounded below and hence it converges for any $x_0.$

Answer 3

There is a nice substitution and technique applied here. It enables us to find a closed form of $x_n$ from which the correct alternative can be easily derived. Given \begin{align*} x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\qquad\qquad n\geq 1 \end{align*} we substitute $\color{blue}{x_n=\sqrt{6}\,y_n}$ and we obtain \begin{align*} y_n&=\frac{y_{n-1}}{2}+\frac{1}{2y_{n-1}}\\ &=\frac{y_{n-1}^2+1}{2y_{n-1}}\tag{1} \end{align*} From (1) we get \begin{align*} y_n-1&=\frac{\left(y_{n-1}-1\right)^2}{2y_{n-1}}\\ y_n+1&=\frac{\left(y_{n-1}+1\right)^2}{2y_{n-1}}\\ \frac{y_n-1}{y_n+1}&=\left(\frac{y_{n-1}-1}{y_{n-1}+1}\right)^2=\cdots=\underbrace{\left(\frac{y_0-1}{y_0+1}\right)^{2^n}}_{C_n}\tag{2} \end{align*} We can now calculate $y_n$ from (2) and obtain a closed form of $x_n$ this way: \begin{align*} y_{n}-1&=C_n\left(y_n+1\right)\\ y_n\left(1-C_n\right)&=1+C_n\\ y_n&=\frac{1+C_n}{1-C_n}\qquad\Rightarrow\qquad \color{blue}{x_n=\sqrt{6}\,\frac{1+\left(\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right)^{2^n}} {1-\left(\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right)^{2^n}}\qquad n\geq 0}\tag{3} \end{align*}

We conclude from (3)

\begin{align*} \left\{x_n\right\}_{n\geq 0}\ \mathrm{convergent}\quad&\Leftrightarrow\quad\left|\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right|<1\\ &\Leftrightarrow\quad\left|x_0-\sqrt{6}\right|<\left|x_0-\left(-\sqrt{6}\right)\right|\\ &\,\,\color{blue}{\Leftrightarrow\quad x_0>0} \end{align*} and (b) is the correct answer.