Q:

Given $x_0>0$ and $a>0$, if $x_{n+1}=\frac{2a^2x_n}{x_n^2 + a^2}$ for every $n\ge0$ then $\lim\limits_{n\to\infty}x_n=a$

Given $x_n$ be a sequence of positive numbers such that 
$$x_{n+1}=\frac{2a^2x_n}{x_n^2 + a^2} , a>0.$$
Show that 
$$\lim_{n\to\infty}x_n=a.$$  
I thought of first proving sequence bounded and then either increasing or decreasing, so that i could establish its convergence. But i couldnot do it. Thanks for helping

A:

Firstly, you have $x_n >0$, $\forall n$, then $x_n^2 + a^2 \geq 2ax_n$.
It implies that that $x_{n+1} \leq \frac{2a^2x_n}{2ax_n} = a$. 
It means that $x_n \leq a$, for all $n \geq 1$ (maybe not $n=0$, i.e $x_0 > a$).
So, you also have $x_{n+1} \geq \frac{2a^2x_n}{2a^2} = x_n$, (because $ 0 < x_n \leq a$).
You can conclude that this sequence is converge. 

A:

$x_1=a\tanh(\theta)\Rightarrow x_2=a\frac{2\tanh(\theta)}{1+\tanh^{2}(\theta)}=a\tanh(2\theta)\Rightarrow ...\Rightarrow x_n=a\tanh(2^{n-1}\theta)$

