Limit of $a_{n+1}=\frac{2a_n^3}{1+a_n^4}$

Question

Let $a_1$ be real, and define $$a_{n+1}=\frac{2a_n^3}{1+a_n^4}$$ How can I prove that this $\{a_n\}$ to have limit.

I find it is hard to track. What I can do is just when $a_1=1$ then $a_n=1$; when $a_1=-1$, then $a_n=-1$; when $|a_1|<1$, $a_n\to 0$. When $|a_1|>1$, I have not find any idea.

Answer 1

We are given the recursion $$a_0:=a>0,\qquad a_{n+1}:=f(a_n)\quad(n\geq0)$$ with $$f(x):={2x^3\over 1+x^4}\ .$$ The following figure shows the graph of $f$ for $x\geq0$:

Since $$f(x)\geq0,\qquad x-f(x)={x(1-x^2)^2\over 1+x^4}\geq0$$ (with strict inequality unless $x=0$ or $x=1$), we deduce that the the sequence $(a_n)_{n\geq0}$ is monotonically decreasing towards a limit $\alpha\geq0$. This limit necessarily satisfies $f(x)=x$, whence $\alpha\in\{0,1\}$, and may depend on the starting value $a$.

It follows that for $0<a<1$ we have $\lim_{n\to\infty} a_n=0$.

When $a=1$ one has $a_n=1$ for all $n\geq0$.

In order to analyze the case $a>1$ we write $f$ in the form $$f(x)=1+{x-1\over1+x^4}(1+x+x^2-x^3)\ .\tag{1}$$ Here the right hand side is $>1$ for $1<x<\xi\doteq1.83929$, where $\xi$ is the real zero of the polynomial on the right of $(1)$, and is $<1$ when $x>\xi$; see also the figure.

It follows that for $1<a<\xi$ we have $\lim_{n\to\infty} a_n=1$. When $a=\xi$ we obtain $a_1=1$ and then $a_n=1$ for all $n\geq1$. When $a>\xi$ then $a_1<1$ and as a consequence $\lim_{n\to\infty} a_n=0$.

Answer 2

It is sufficient to consider $a_n > 0$, as otherwise you could do the same analysis with $b_n = -a_n$ and $a_1=0 \iff a_n=0$, so that's trivial.

Note that $1+x^4 \ge 2x^2$ by AM-GM with equality iff $x=1$, so we have $a_{n+1} < a_n$, unless $a_k=1 \implies a_{n> k}=1$. Thus we have a decreasing, positive (i.e. bounded) sequence ...