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.
Best Answer
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 $x_0>0$.