Suppose $m \geq 2$ be an integer such that $m \geq 2$, let $a, x_0 >0$. The sequence is then
$x_{n+1} = \frac{1}{m}\left( (m-1)x_n + \frac{a}{x^2 _{m+1}} \right )$
Show that the sequence $(x_n)_{n \in \mathbb{N}}$ converges to a unique positive solution of $x^m = a$
Attempt
Honestly, I thought I would just go and prove this without the whole m getting in my way first. So I would try to prove by m = 2 first and then use the fact that $m \geq 2$ to cheat my way through it.
But when I got to the limit part to converging to $x^m = a$, I couldn't get it. I got nothing like it.
EDIT
Sorry, it should have been
$x_{n+1} = \frac{1}{m}\left( (m-1)x_n + \frac{a}{x^{m-1}_n} \right )$
@ Gerry
When $m = 2$
$x_{n+1} = \frac{1}{2}\left( (2-1)x_n + \frac{a}{x^{2-1}_n} \right ) = \frac{1}{2}\left( (x_n + \frac{a}{x_n}) \right )$
EDIT2: yes it worked for $m = 2$