Does $x_{n+1}=\frac{x_n+\alpha}{x_n+1}$ converge? Problem
Let $\alpha > 1$. A sequence $(x_n)_{n \ge 0} $ is defined such that $x_0 > \sqrt{\alpha}$ and 
$$x_{n+1}=\frac{x_n+\alpha}{x_n+1} \ ,\ \forall n \in \mathbb{N^0}$$
Does this sequence converge? If so, to what?
I solved the sum in a way and so want someone to check its validity. If there is error please correct it. And also if there is any other alternate solution please show it. I have attached a link to my solution:
My Solution
 A: First, note that if this sequence does converge, the limit $ x $ satisfies
$$ x(x+1) = x + \alpha $$
$$ x^2 - \alpha = 0 $$
so that if the sequence converges, it can only converge to $ \sqrt{\alpha} $. (The terms are nonnegative.) Now, we show that it converges. Note that
$$ |x_{n+1} - x_n| = \left| \frac{(x_n + \alpha)(x_{n-1} + 1) - (x_{n-1} + \alpha)(x_n + 1)}{(x_n + 1)(x_{n-1} + 1)} \right| = \left| \frac{(x_n - x_{n-1})(1 - \alpha)}{(x_n + 1)(x_{n-1} + 1)} \right| $$
Now, we have the following bound:
$$ (x_n + 1)(x_{n-1} + 1) = x_{n-1} + \alpha + x_{n-1} + 1 > \alpha + 1 $$
therefore
$$ |x_{n+1} - x_n| < \left| \frac{\alpha - 1}{\alpha + 1} \right| |x_n - x_{n-1}| $$
This establishes that $ x_n $ is a contractive sequence for $ \alpha > 1 $, therefore it is convergent.
A: If the limit $L$ exist, it will be such that $$L=\frac{L+\alpha}{L+1}$$ Reducing to same denominator $$L^2+L=L+\alpha$$ I am sure that you can take it from here.
A: Since $\alpha\gt1$, we get
$$
\begin{align}
x_{n+1}
&=\frac{x_n+\alpha}{x_n+1}\\
&=\frac{\alpha-1}{x_n+1}+1\\[6pt]
&\gt1\tag{1}
\end{align}
$$
Iterating the recursion, we get
$$
\begin{align}
x_{n+2}
&=\frac{\frac{x_n+\alpha}{x_n+1}+\alpha}{\frac{x_n+\alpha}{x_n+1}+1}\\
&=\frac{x_n+\alpha+\alpha(x_n+1)}{x_n+\alpha+x_n+1}\\
&=\frac{x_n(\alpha+1)+2\alpha}{2x_n+(\alpha+1)}\tag{2}
\end{align}
$$
Therefore, since $x_n\gt1$, if we subtract $\sqrt\alpha$ from $(2)$, we get
$$
\begin{align}
\left|\,x_{n+2}-\sqrt\alpha\,\right|
&=\left|\,\frac{x_n(\alpha-2\sqrt\alpha+1)-\sqrt\alpha(\alpha-2\sqrt\alpha+1)}{2x_n+(\alpha+1)}\,\right|\\[3pt]
&=\left|\,x_n-\sqrt\alpha\,\right|\frac{(\alpha+1)-2\sqrt\alpha}{(\alpha+1)+2x_n}\\[3pt]
&\le\left|\,x_n-\sqrt\alpha\,\right|\color{#C00000}{\frac{(\alpha+1)-2\sqrt\alpha}{(\alpha+1)+2}}\tag{3}
\end{align}
$$
Since $0\lt\frac{\left(\sqrt\alpha-1\right)^2}{\alpha+3}=\color{#C00000}{\frac{(\alpha+1)-2\sqrt\alpha}{(\alpha+1)+2}}=1-2\frac{\sqrt\alpha+1}{\alpha+3}\lt1$,
inequality $(3)$ says that both even and odd indices converge to $\sqrt\alpha$.
A: This recurrence relation looked familiar, and sure enough I showed how to solve this class of problem earlier. First we rewrite as
$$x_{n+1}=\frac{x_n+\alpha}{x_n+1}=1+\frac{\alpha-1}{x_n+1}$$
So now we let $y_n=x_n-1$ and we have
$$y_{n+1}=\frac{\alpha-1}{y_n+2}$$
Now the trick that works is
$$z_n=\frac1{y_n+c}$$
Where $c$ is to be determined. Then $y_n=1/z_n-c$ and
$$\frac1{z_{n+1}}-c=\frac{\alpha-1}{\frac1{z_n-c+2}}=\frac{1-cz_{n+1}}{z_{n+1}}=\frac{(\alpha-1)z_n}{1+(2-c)z_n}$$
Clearing denominators,
$$(\alpha-1)z_nz_{n+1}=1-cz_{n+1}+(2-c)z_n-c(2-c)z_nz_{n+1}$$
We can eliminate the nonlinear term if $c^2-2c-\alpha+1=0$, so $c=1\pm\sqrt{1+\alpha-1}=1-\sqrt{\alpha}$ so as to make $z_n\rightarrow\infty$ as $n\rightarrow\infty$. Then we have a linear difference equation with constant coefficients,
$$(1-\sqrt{\alpha})z_{n+1}-(1+\sqrt{\alpha})z_n=1$$
The homogeneous difference equation is
$$(1-\sqrt{\alpha})z_{h,n+1}-(1+\sqrt{\alpha})z_{h,n}=0$$
Letting $z_{h,n}=r^n$ we derive the characteristic equation
$$(1-\sqrt{\alpha})r-(1+\sqrt{\alpha})=0$$
So the solution to the homogeneous equation is
$$z_{h,n}=K\left(\frac{1+\sqrt{\alpha}}{1-\sqrt{\alpha}}\right)^2$$
And trying a constant solution we find soon enough that
$$z_{p,n}=\frac{-1}{2\sqrt{\alpha}}$$
is a particular solution to the inhomogeneous difference equation. Thus the general solution is
$$z_n=z_{p,n}+z_{h,n}=\frac{-1}{2\sqrt{\alpha}}+K\left(\frac{1+\sqrt{\alpha}}{1-\sqrt{\alpha}}\right)^2$$
Applying initial conditions,
$$z_0=\frac{-1}{2\sqrt{\alpha}}+K=\frac1{y_0+1-\sqrt{\alpha}}=\frac1{x_0-\sqrt{\alpha}}$$
So
$$K=\frac1{x_0-\sqrt{\alpha}}+\frac1{2\sqrt{\alpha}}=\frac{x_0+\sqrt{\alpha}}{2\sqrt{\alpha}\left(x_0-\sqrt{\alpha}\right)}$$
$$z_n=\frac{-1}{2\sqrt{\alpha}}+K\left(\frac{1+\sqrt{\alpha}}{1-\sqrt{\alpha}}\right)^n$$
Working back to the start,
$$x_n=\frac1{\frac{-1}{2\sqrt{\alpha}}+K\left(\frac{1+\sqrt{\alpha}}{1-\sqrt{\alpha}}\right)^n}+\sqrt{\alpha}$$
Since $\alpha>1$, it follows that
$$\lim_{n\rightarrow\infty}\left|\frac{1+\sqrt{\alpha}}{1-\sqrt{\alpha}}\right|^n=\infty$$
So
$$\lim_{n\rightarrow\infty}x_n=\sqrt{\alpha}$$
Perhaps a little more long-winded, but we were able to derive an expression for $x_n$.
