Convergence of $(x_n)$ defined by $x_1=2$ and $x_{n+1} = \frac13(2x_n + {3\over\ x_n^2})$ The next sequence is defined by recursion
$x_1=2$ and $x_{n+1} = \frac13(2x_n + {\frac3{x_n^2})}$
Show that the sequence is convergent.
My solution is
$|x_{n+2}-x_{n+1}|=|{{2x_{n+1} + {3\over\ x_{n+1}^2}}\over\ 3}- {{2x_n + {3\over\ x_n^2}}\over\ 3}|\\ < {2|x_{n+1}-x_{n}|\over\ 3}+ 
|{{1\over\ x_{n+1}^2}-{1\over\ x_{n}^2}}|<{2|x_{n+1}-x_{n}|\over\ 3}$
then is a sequence of cauchy, then is convergent,
Is it correct?
 A: Take the limit as n approaches infinity to see if the series converge or not:
$$ \lim_{n \rightarrow\infty}\frac{2n^{3}+3}{3n}$$
$$ \lim_{n \rightarrow\infty}\frac{n^{3}(2+\frac{3}{n^{3}})}{3n}$$
$$ \lim_{n \rightarrow\infty}\frac{2n^{3}}{3n} $$
$$ \lim_{n \rightarrow\infty}\frac{2n^{2}}{3}=\infty $$
The series diverge.
A: Now my solution is 
$x_{n+1}^3>3$ for induction 
$x_{1}^3=8>3$
and if $x_{n}^3>3$ 
then 
$x_{n+1}^3= \frac{({2x_n+{3\over \ x_n^2}})^3}{27}
={8x_n^3({1+{3\over \ 2x_n^3}})^3 \over \ 27}>{8({1+{9\over \ 2x_n^3}}) }>8>3$
and ${x_{n+1}\over\ x_{n}} = {2x_n+ {3\over \ x_n^2}\over \ 3x_n} = {2+ {3\over \ x_n^3}\over \ 3}<{2+ {3\over \ 3}\over \ 3}=1$
then the sequence its convergence
A: The problem is closely related to Newton's method for finding numerical solutions of the equation $f(x)=0$ with $f(x)=x^3-c$, $c\in\mathbb R$. The sequence
$$
x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^3-c}{3x_n^2}=\frac{2x_n^3-c}{3x_n^2}=\frac 13\left(2x_n-\frac c{x_n^2}\right)
$$
converges to the root of the equation $x^3=c$. It is enough to set $c=3$.
