Convergence of a sequence of real numbers Let $\alpha, \gamma$ be real numbers such that $0<\alpha<1$ and $\gamma>0$. Consider the sequence of real numbers given by
$$
\begin{cases}
x_0\ne 0&\\
x_{k+1}=x_k\left(1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right) \quad (k\in \mathbb{N}).&
\end{cases}
$$
Suppose that $x_k\ne 0$ for all $k\in \mathbb{N}$.
Prove that :


*

*The sequence $\{x_k\}_{k\in\mathbb{N}}$ does not converge.

*The sequence $\{|x_k|\}_{k\in\mathbb{N}}$ converges to $[(1/2)(1+\alpha)\gamma]^{1/(1-\alpha)}.$
 A: Let $L=(\gamma(1+\alpha)/2)^{1/(1-\alpha)}$.
Suppose that $|x_k| < L = (\gamma(1+\alpha)/2)^{1/(1-\alpha)}=((\gamma(1+\alpha)/2)^{\alpha/(1-\alpha)})^{1/\alpha}=((\gamma(1+\alpha)/2)^{-1}(\gamma(1+\alpha)/2)^{1/(1-\alpha)})^{1/\alpha}=((\gamma(1+\alpha)/2)^{-1}L)^{1/\alpha}$
from which we conclude that $|x_k|^{\alpha}(\gamma(1+\alpha)) < 2L$. Also $|x_k| < L=(\gamma(1+\alpha)/2)^{1/(1-\alpha)}$ implies that $\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}>2$.
Then $|x_{k+1}|=|x_k|\left|1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right|= |x_k|\left(\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}-1\right)>|x_k|$.
If $|x_{k+1}| >L$ then from $|x_k|^{\alpha}(\gamma(1+\alpha)) < 2L$ we manipulate to get
$|x_k|\left(\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}-1\right) -L< L-|x_k|$
and finally $|x_k|\left|1-\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}\right| -L< L-|x_k|$. So we conclude that $|x_{k+1}|-L < L - |x_k|$ and then from $|x_{k+1}| >L$  and $|x_k| < L$ we know that $||x_{k+1}|-L| < ||x_k|-L|$.
Supppose that $|x_k| > L = (\gamma(1+\alpha)/2)^{1/(1-\alpha)}=((\gamma(1+\alpha)/2)^{-1}L)^{1/\alpha}$
then $|x_k|^{\alpha}(\gamma(1+\alpha)) > 2L$.
Also $0<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<2$ and
$|x_{k+1}|=|x_k|\left|1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right|<|x_k|$.
If $|x_{k+1}| < L$ and $1<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<2$ then from $|x_k|^{\alpha}(\gamma(1+\alpha)) > 2L$ we manipulate to get
$|x_k|\left(\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}-1\right) -L> L-|x_k|$
and finally $|x_k|\left|1-\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}\right| -L> L-|x_k|$. So we conclude that $|x_{k+1}|-L > L - |x_k|$ and $L-|x_{k+1}| > |x_k|-L$ then from $|x_{k+1}| <L$  and $|x_k| > L$ we know that $||x_{k+1}|-L| < ||x_k|-L|$. If $|x_{k+1}| <L$ and $0<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<1$ then $\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<1$ implies that $|x_k| > 2^{1/(1-\alpha)}L\ge 2L$. And from $|x_{k+1}|<L$ we know $||x_{k+1}|-L|<L< |x_k| - L = ||x_k| - L|$.
In summary we have shown that if $|x_k| < L$ then $|x_{k+1}| >|x_k|$. If $|x_k| < L$ and $|x_{k+1}| <L$ then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| < L$ and $|x_{k+1}| >L$ then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| > L$ then $|x_{k+1}| <|x_k|$. If $|x_k| > L$ and $|x_{k+1}| >L$ then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| > L$ and $|x_{k+1}| <L$ then $||x_{k+1}|-L|<||x_k|-L|$. No matter what the case it is true that $||x_{k+1}|-L|<||x_k|-L|$. Therefore the sequence $|x_k|-L$ converges. Showing that the limit of $|x_k|$ must be $L$ is easy.
A: We only need to prove the second claim for the following reason.
If $|x_k|\rightarrow (\gamma (1+\alpha)/2)^{1/(1-\alpha)}$, 
$$\frac{x_{k+1}}{x_k}=\left(1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right) \rightarrow-1.$$ 
Thus ${x_{k}}$ will be an oscillating sequence.
I will write it later about how to prove the second claim.
