Convergence of a sequence of real numbers Let $\alpha, \gamma$ be real numbers such that $0&lt;\alpha&lt;1$ and $\gamma&gt;0$. Consider the sequence of real numbers given by
$$
\begin{cases}
x_0\ne 0&amp;\\
x_{k+1}=x_k\left(1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right) \quad (k\in \mathbb{N}).&amp;
\end{cases}
$$
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 $[(1/2)(1+\alpha)\gamma]^{1/(1-\alpha)}.$
Let $L=(\gamma(1+\alpha)/2)^{1/(1-\alpha)}$.

Suppose that $|x_k| &lt; L = (\gamma(1+\alpha)/2)^{1/(1-\alpha)}=((\gamma(1+\alpha)/2)^{\alpha/(1-\alpha)})^{1/\alpha}=((\gamma(1+\alpha)/2)^{-1}(\gamma(1+\alpha)/2)^{1/(1-\alpha)})^{1/\alpha}=((\gamma(1+\alpha)/2)^{-1}L)^{1/\alpha}$

from which we conclude that $|x_k|^{\alpha}(\gamma(1+\alpha)) &lt; 2L$. Also $|x_k| &lt; L=(\gamma(1+\alpha)/2)^{1/(1-\alpha)}$ implies that $\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}&gt;2$.
Then $|x_{k+1}|=|x_k|\left|1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right|= |x_k|\left(\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}-1\right)&gt;|x_k|$.
If $|x_{k+1}| &gt;L$ then from $|x_k|^{\alpha}(\gamma(1+\alpha)) &lt; 2L$ we manipulate to get
$|x_k|\left(\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}-1\right) -L&lt; L-|x_k|$
and finally $|x_k|\left|1-\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}\right| -L&lt; L-|x_k|$. So we conclude that $|x_{k+1}|-L &lt; L - |x_k|$ and then from $|x_{k+1}| &gt;L$  and $|x_k| &lt; L$ we know that $||x_{k+1}|-L| &lt; ||x_k|-L|$.

Supppose that $|x_k| &gt; L = (\gamma(1+\alpha)/2)^{1/(1-\alpha)}=((\gamma(1+\alpha)/2)^{-1}L)^{1/\alpha}$
then $|x_k|^{\alpha}(\gamma(1+\alpha)) &gt; 2L$.
Also $0&lt;\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}&lt;2$ and
$|x_{k+1}|=|x_k|\left|1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right|&lt;|x_k|$.
If $|x_{k+1}| &lt; L$ and $1&lt;\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}&lt;2$ then from $|x_k|^{\alpha}(\gamma(1+\alpha)) &gt; 2L$ we manipulate to get
$|x_k|\left(\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}-1\right) -L&gt; L-|x_k|$
and finally $|x_k|\left|1-\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}\right| -L&gt; L-|x_k|$. So we conclude that $|x_{k+1}|-L &gt; L - |x_k|$ and $L-|x_{k+1}| &gt; |x_k|-L$ then from $|x_{k+1}| &lt;L$  and $|x_k| &gt; L$ we know that $||x_{k+1}|-L| &lt; ||x_k|-L|$. If $|x_{k+1}| &lt;L$ and $0&lt;\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}&lt;1$ then $\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}&lt;1$ implies that $|x_k| &gt; 2^{1/(1-\alpha)}L\ge 2L$. And from $|x_{k+1}|&lt;L$ we know $||x_{k+1}|-L|&lt;L&lt; |x_k| - L = ||x_k| - L|$.

In summary we have shown that if $|x_k| &lt; L$ then $|x_{k+1}| &gt;|x_k|$. If $|x_k| &lt; L$ and $|x_{k+1}| &lt;L$ then $||x_{k+1}|-L|&lt;||x_k|-L|$. If $|x_k| &lt; L$ and $|x_{k+1}| &gt;L$ then $||x_{k+1}|-L|&lt;||x_k|-L|$. If $|x_k| &gt; L$ then $|x_{k+1}| &lt;|x_k|$. If $|x_k| &gt; L$ and $|x_{k+1}| &gt;L$ then $||x_{k+1}|-L|&lt;||x_k|-L|$. If $|x_k| &gt; L$ and $|x_{k+1}| &lt;L$ then $||x_{k+1}|-L|&lt;||x_k|-L|$. No matter what the case it is true that $||x_{k+1}|-L|&lt;||x_k|-L|$. Therefore the sequence $|x_k|-L$ converges. Showing that the limit of $|x_k|$ must be $L$ is easy.
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.