The sequence was defined by the equations: $${a}_{1}=a\left(\in R \right),{a}_{n+1}=\frac{2{{a}_{n}}^{3}}{1+{{a}_{n}}^{4}},n\geq 1.$$
show that
$\left(a \right)$The given sequence is convergent. (whatever $a\in \mathbf{R}$)
$\left(b \right)$ Finding its all possible limit.
$\left(c \right)$ Dividing $\mathbf{R}$ into several intervals,such that if the initial value ${a}_{1}$ lies in the same interval,then sequence $\begin{Bmatrix} {a}_{n} \end{Bmatrix}$ has the same limit.
On
Hint: Finding all possible possible limits for part b) is very simple. Suppose one possible limit of the sequence is $L=\lim_n a_n$. Taking the limits of both sides of the recurrence relation you find:
$$L=\frac{2L^3}{1+L^4}.$$
That is, if any limits exist then they must satisfy the above equation.
*Note: just because a real number satisfies the above equation doesn't guarantee there actually does exist a sequence converging to that limit, so be careful not to over think part c).
On
When $a=0$ convergence is trivial.
Note that $1+(a_n)^4\ge 2(a_n)^2$(Because this reduces to $(a_n^2-1)^2\ge 0$), hence $a_{n+1}=\frac{2(a_n)^3}{1+{a_n}^4}\le a_n$. If $a>0$ then the sequence if positive and zero is a lower bound and monotone-convergence theorem says this converges, If $a<0$, using the same argument you can prove that $-a_n$ converges and hence $a_n$ converges. Hence $\lim\limits_{n\rightarrow \infty}=L$
Now to calculate the limit, all we need to solve is $L=\frac{2(L)^3}{1+{L}^4}$ and that gives $L=0,1,-1$. When $a=0$ it converges to $L=0$. We also see that: If $a>1$ then $a_2<1$ hence when $a>0$ then $a_n$ converges to $0$ when $a=1$ it is also clear that $a_n$ converges to $1$. As the positive part is clear, for negative part all you have to do is considering $-a_n$.
On
Let ${x}_{0}$ is the positive root of equation ${x}^{3}-{x}^{2}-x-1=0$.
Some thoughts on (a)
If $a>0$ use the AM-GM inequality to show that
$a_{n+1}\leq a_{n}$ for all n.
In this case we also have
$a_{n}\geq 0$ for all n.
What is that telling you about convergence?