If $\alpha$ $\in$ $(0,2)$ study the convergence of the sequence given with recurrence relation $X_{n+1}=\alpha X_{n}-(1-\alpha)X_{n-1}$ If $\alpha$ $\in$ $(0,2)$ study the convergence of the sequence given with recurrence relation $X_{n+1}=\alpha X_{n}-(1-\alpha)X_{n-1}$.
Find the limit of the sequence.
Can somebody help me with this problem?
I only studied sequences with recurrence relation give with firsts conditions. I don't know how to deal with these ones.
I would appreciate some help

$X_n=Aa^n+Bb^n$ for some constants $A$ and $B$, where $a$ and $b$ are the roots of the equation $x^2-\alpha x+(1-\alpha)=0$. The sequence is convergent if both $|a|\le 1$ and $|b|\le 1$.
When $\alpha^2-4(1-\alpha)<0$, i.e. when $0<\alpha<2\sqrt{2}-2$, the quadratic equation has unreal roots, $|a|=|b|$ and $|a||b|=|ab|=1-\alpha<1$. The sequence is convergent.
When $\alpha^2-4(1-\alpha)\ge0$, i.e. when $2\sqrt{2}-2\le \alpha<2$, the quadratic equation has real roots. Let $a\le b$ Then $\displaystyle a=\frac{\alpha-\sqrt{\alpha^2+4\alpha-4}}{2}=\frac{2(1-\alpha)}{\alpha+\sqrt{\alpha^2+4\alpha-4}}$ and $\displaystyle b=\frac{\alpha+\sqrt{\alpha^2+4\alpha-4}}{2}$.
So, when $2\sqrt{2}-2\le \alpha\le 1$, $0\le a\le b$. We have to consider $|b|$ only.
\begin{align*}
|b|=\frac{\alpha+\sqrt{\alpha^2+4(\alpha-1)}}{2}\le \frac{\alpha+\sqrt{\alpha^2+4(\alpha-\alpha)}}{2}=\alpha\le1
\end{align*}
The sequence is convergent.
If $1<\alpha<2$, then
\begin{align*}
|b|=\frac{\alpha+\sqrt{\alpha^2+4(\alpha-1)}}{2}\ge \frac{\alpha+\sqrt{\alpha^2}}{2}=\alpha>1
\end{align*}
The sequence is divergent.

The sequence is convergent when $0<\alpha\le1$.
When $0<\alpha<1$, $\displaystyle \lim_{n\to\infty}X_n=0$.
When $\alpha=1$, $X_{n+1}=X_n$ and the seqeunce is a constant sequence. 

As Youem pointed out in the comment, there is a special case that the sequence is convergent even when $1\le\alpha<2$.
If $1\le\alpha<2$, then
\begin{align*}
0\ge\frac{2(1-\alpha)}{\alpha+\sqrt{\alpha^2+4\alpha-4}}\ge\frac{2(1-\alpha)}{\alpha+\sqrt{\alpha^2+4\alpha-4\alpha}}=\frac{1}{\alpha}-1\ge-1
\end{align*}
and hence $|a|\le1$.
$X_n=Aa^n+Bb^n$ is convergent (when $1<\alpha<2$) if $B=0$, i.e. $X_n=Aa^n$
So, when  $1\le\alpha<2$, The sequence is convergent to $0$ when $\displaystyle X_2=\left(\frac{\alpha-\sqrt{\alpha^2+4\alpha-4}}{2}\right)X_1$.
