<s>[INST] Find $\lim_{n\to\infty} \lvert x_n \rvert ^\frac{1}{n}$ when $x_{n+1}=3x_n + \frac{2}{x_n^2},x_1=1$ $\lim_{n\to\infty} \lvert x_n \rvert ^\frac{1}{n}$ when $x_{n+1}=3x_n + \frac{2}{x_n^2}, x_1=1$
I figured out that $x_n$ increases as n approaches infinity.
However, the power approaches 0 as n approaches infinity.
So, I thought the limit does not approach 0 because the base approaches infinity while the power approaches 0. 
Any hint that I can get?
 [/INST] Assuming that
$x_0 > 0$,
$x_{n+1}
=3x_n + \frac{2}{x_n^2}
\gt 3x_n
$
so
$x_n > 3^n x_0$.
Also
$x_{n+1}
=3x_n + \frac{2}{x_n^2}
\lt 3x_n+\frac{2}{(3^nx_0)^2}
= 3x_n+\frac{2}{9^nx_0^2}
$
so
$\dfrac{x_{n+1}}{3^{n+1}}
\lt \dfrac{x_n}{3^n}+\frac{2}{27^nx_0^2}
$.
Letting
$y_n = \dfrac{x_n}{3^n}$,
$y_{n+1}-y_n
\lt \frac{2}{27^nx_0^2}
$.
Summing
$y_m-y_0
=\sum_{n=0}^{m-1}(y_{n+1}-y_n)
\lt \sum_{n=0}^{m-1} \frac{2}{27^nx_0^2} 
=\frac{2}{x_0^2}  \sum_{n=0}^{m-1} \frac{1}{27^n} 
\lt \frac{2}{x_0^2}\dfrac{1}{1-1/27}
= \frac{27}{13x_0^2}
=c$
so
$y_m
\lt y_0+c
$
or
$\dfrac{x_m}{3^m}
\lt y_0+c
$
or
$\dfrac{x_m^{1/m}}{3}
\lt (y_0+c)^{1/m}
\to 1
$.
Therefore
$\lim_{m \to \infty} x_m^{1/m}
=3
$.
</s>