Converging Sequences How can I show that the sequence $\{x_n\}_{n=1}^{\infty}$ defined by $$x_{n+1} = \frac{x_n(x_{n}^2 + 3a)}{3x_{n}^2 + a}$$ is convergent? I don't think plugging this into the convergent-sequences definition will help and i've been stuck on this for awhile. Someone suggested cauchy criterion, but I don't know how to apply that. 
Apologies, I forgot to include that $a>0$, $x_1 > 0$.
 A: Working from the defining equation
$$
x_{n+1}=\frac{x_n(x_n^2+3a)}{3x_n^2+a}\tag{1}
$$
we have
$$
x_{n+1}=\frac{2x_n(\sqrt{a}+x_n)}{3x_n^2+a}\sqrt{a}+\frac{(x_n-\sqrt{a})^2}{3x_n^2+a}x_n\tag{2}
$$
and
$$
\frac{2x_n(\sqrt{a}+x_n)}{3x_n^2+a}+\frac{(x_n-\sqrt{a})^2}{3x_n^2+a}=1\tag{3}
$$
Since both $\frac{2x_n(\sqrt{a}+x_n)}{3x_n^2+a}$ and $\frac{(x_n-\sqrt{a})^2}{3x_n^2+a}$ are at least $0$, Equations $(2)$ and $(3)$ say that $x_{n+1}$ is between $x_n$ and $\sqrt{a}$ (inclusive). This is enough to guarantee convergence (if $x_1\le\sqrt{a}$, then $x_n$ is non-decreasing and bounded above by $\sqrt{a}$; if $x_1\ge\sqrt{a}$, then $x_n$ is non-increasing and bounded below by $\sqrt{a}$).
To find the limit, subtract $x_n$ from $(2)$ and take the limit as $n\to\infty$:
$$
0=\frac{2x_\infty(\sqrt{a}+x_\infty)}{3x_\infty^2+a}(\sqrt{a}-x_\infty)\tag{4}
$$
Since $x_\infty$ is between $x_1$ and $\sqrt{a}$ (inclusive), it must be greater than $0$. Therefore, we must have $x_\infty=\sqrt{a}$.
A: Let
$$
f(x)=\frac{x(x^2+a)}{3\,x^2+a},\quad x\ge0.
$$
$f(x)=x$ has two non-negative solutions: $0$ and $\sqrt a$. Moreover
$$
x<f(x)<\sqrt a\quad\text{if}\quad 0<x<\sqrt a,
$$
and
$$
\sqrt a<f(x)<x\quad\text{if}\quad x>\sqrt a.
$$



*

*If $x=\sqrt a$ then $x_n=\sqrt a$ for all $n$.

*If $0<x_1<\sqrt a$, then $x_n$ is increasing and bounded above by $\sqrt a$. This implies that it converges, and the limit must be $\sqrt a$.

*If $x_1>\sqrt a$, then $x_n$ is decreasing and bounded below by $\sqrt a$. This implies that it converges, and the limit must be $\sqrt a$.

A: I tried this for generic $a$ aand $x_0$, and the answer is that it does not converge for all values of $x_0$ if $a < 0$.
A tempting first transformation is to scale out $a$ by substituting $u = \frac{x^2}{a}$ giving
$$
u_{n+1} = u_n\left( \frac{u_n+3}{3u_n+1} \right)^2
$$
We immediately see that this converges for $u_0 \geq 0$, but when $u_0 < 0$ it goes chatic for most values of $u_0 < 0$, and has some 2-cycles which are unstable (for example, start with $$ u_0 = -0.527864045$$
which stays near an oscillation alternating with $-9.472136$ for about 10 oscilations before moving far enough away to get chaotic; there is a fixed point near there.
