Is the sequence $x_{n+1} = \frac{x_n + \alpha}{x_n + 1}$ convergent?

Question

Fix $\alpha > 1$, and consider the sequence $(x_n)_n \geq 0$ defined by $x_0 > \sqrt \alpha$ and $$x_{n+1} = \frac{x_n + \alpha}{x_n + 1}, n = 0, 1, 2, \dots$$ Does this sequence converge, and if, to what?

I tried to get the difference between $x_n$ and $x_{n+1}$, it didn't work out. How do I go about solving it?

Answer 1

Consider an auxiliary sequence $y_n=\dfrac{x_n-\sqrt{\alpha}}{x_n+\sqrt{\alpha}}$, this is well-defined since a simple induction shows that $x_n\geq 1$ for all $n$. Now, using the definition we see that $$ \forall\,n\geq0,\qquad y_{n+1}=-\frac{\sqrt\alpha-1}{\sqrt\alpha+1}y_n $$ Thus, setting $\lambda=\frac{\sqrt\alpha-1}{\sqrt\alpha+1}\in(0,1)$, we get $$ \forall\,n\geq0,\qquad y_n=(-1)^n\lambda^ny_0 $$ So $\lim_{n\to\infty}y_n=0$, and consequently $\lim_{n\to\infty}x_n=\sqrt\alpha$, because $x_n-\sqrt\alpha=\dfrac{2\sqrt\alpha y_n}{1-y_n}$.

Answer 2

This is my first answer, so if anything appear unclear feel free to write to me :)

Well.. What you are presenting, is a nonlinear inhomogeneous difference equation. (at least that's what I learned it by).

Now first assume that $\{x_n\}$ converge against a value $x$. Then:

$\text{lim}_{n\rightarrow\infty}(x_{n+1})=\text{lim}_{n\rightarrow\infty}\left(\frac{x_n+\alpha}{x_n+1}\right)$

$x = \frac{x+\alpha}{x+1}$

$ x^2-\alpha=0$

$ x=\pm\sqrt{\alpha}$

So if $\{x_n\}$ converges, it converges against either $\sqrt{\alpha}$ or $-\sqrt{\alpha}$.

Now lets examine if $\{x_n\}$ is decreasing. If its decreasing, it meets the condition:

$x_{n+1}<x_n$

$\frac{x_{n}+\alpha}{x_n+1}<x_n$

$\pm\sqrt{\alpha}<x_n$

I.e. $\{x_n\}$ is decreasing if and only if $\sqrt{\alpha}<x_n$.

We are now going to show by induction that $\sqrt{\alpha}<x_n$ is in fact true and then it follows that $\{x_n\}$ converges against $\sqrt{\alpha}$. This is due to the fact that $\{x_n\}$ is decreasing and lower bounded. Since $\sqrt{\alpha}<x_n$ it will have to converge against $\sqrt{\alpha}$ and not $-\sqrt{\alpha}$. But now for the induction proof:

We already know that $x_0>\sqrt{\alpha}$. Now assume that $x_n>\sqrt{\alpha}$. Then:

$x_{n+1}=\frac{x_n+\alpha}{x_n+1}<\frac{x_n+x_n^2}{x_n+1}=\frac{x_n(x_n+1)}{x_n+1}<\sqrt{\alpha}$