Linear recursion with convergent coefficients Suppose we have a linear recurrence relation
$$x_n = \alpha_n x_{n-1}+ \beta_n x_{n-2} + y_n,$$
where

*

*$(\alpha_n) \to \alpha >0$,

*$(\beta_n) \to \beta >0$,

*$(y_n) \to y > 0$, and

*the limiting auxiliary equation, $m^2 - \alpha m - \beta$, has roots in $(-1,1)$
(all limits are as $n \to \infty$).
My question:

Does this imply that the sequence $(x_n)_{n=1}^\infty$ converges?

This comes from a convergent example. I am simply wondering if the above information suffices to prove it, or if one needs control on the rates of convergence, say.
 A: The answer to your question is YES ; in fact, I show below that $(x_n)$ always converges to $l=\frac{y}{1-(\alpha+\beta)}$ (notice that $1-(\alpha+\beta)\neq 0$, otherwise $1$ would be a root of $m^2-\alpha m -\beta$).
First, let us see what happens when we adapt the method in the answer to the linked question.
The scalar recursion of order 2 is equivalent to the following recursion
of order 1 for vectors via $v_n=(x_n,x_{n+1})^T$.
\begin{equation}\tag{1}\label{one}
v_{n+1}=A_nv_n+C_n,\ \ A_n=\left(\begin{matrix}0&1\\ \beta_{n+2}& \alpha_{n+2}\end{matrix}\right)
\ C_n=\left(\begin{matrix}0\\y_{n+2}\end{matrix}\right)
\end{equation}
If $\lambda,\mu$ are the roots of $m^2-\alpha m-\beta$ which satisfy $|\lambda|,|\mu|<1$ by assumption  then there is a constant matrix $T$ such that
$$\newcommand{\eps}{\varepsilon}T^{-1}A T= \left(\begin{matrix}\lambda&\eps\\0&\mu\end{matrix}\right)=:D
\mbox{ for }A=\lim A_n=\left(\begin{matrix}0&1\\ \beta & \alpha \end{matrix}\right)$$
where $\eps=0$ unless $\lambda=\mu$  and the eigenvalue $\lambda$ of $A$ has algebraic
multiplicity 1; in this case $\eps\neq0$ and it can be chosen such that
$|\lambda|+|\eps|<1$.
Putting $v_n=Tw_n$ transforms (\ref{one}) to
$$w_{n+1}=(D+F_n)w_n+ C'_n,$$
where $F_n$ is a matrix of elements tending to 0 and $C'_n=T^{-1}C_n$ is a column matrix tending to a finite limit when $n\to \infty$.
Using the maximum norm for vectors and the corresponding matrix norm, we find
$$||w_{n+1}||\leq\big(\max(|\lambda|+|\eps|,|\mu|)+||F_n||\big)||w_n||+||C'_n||.$$
By assumption, the maximum is smaller than 1 and $||F_n||$ tends to 0. Also $(||C'_n||)$ is bounded. Therefore there
exist some $0<M<1$, some $C\gt 0$ and some integer $N$ such that the big parenthesis is $\leq M$, and $||C'_n|| \leq C$ for $n\geq N$. This means that
$$||w_{n+1}||\leq M||w_{n}||+C\mbox{ for }n\geq N$$
and hence $||w_n||\leq ||w_N|| M^{n-N} + C(1+M+M^2+\ldots+M^{n-N}) \leq ||w_N|| M^{n-N} + \frac{C}{1-M}$ for $n\geq N$ :
$$
||w_n||\leq ||w_N|| M^{n-N} + \frac{C}{1-M} \tag{2}\label{two}
$$
Therefore $w_n$ is bounded and therefore also $v_n$ and $x_n$.
Now, if we put $x'_n=x_n-l$, we have
$$
x'_{n}=\alpha_n x'_{n-1}+ \beta_n x'_{n-2} + y'_n
$$
where $y'_n=y_n-(1-\alpha_n-\beta_n)l$ tends to zero. Redoing all our preceding computations with $(x'_n)$ in place of $(x_n)$, we obtain a sequence $||w'_n||$ satisfying an analogue of $\eqref{2}$ where the constant $C$ can be arbitrarily small when $N$ is large enough. This shows that $(w'_n)$ converges to zero, and hence that $(x_n)$ converges to $l$.
A: There is a nice order reduction result for difference inequalities in the following reference which simplifies the reasoning quite a bit:
https://msp.org/involve/2008/1-1/p07.xhtml
For a given $\epsilon > 0$, for sufficiently large N, we have that for $n>N$:
$$(\alpha + \beta -\epsilon)min(x_{n-1}, x_{n-2}) + (\gamma - \epsilon) \leq x_{n} \leq (\alpha + \beta +\epsilon)max(x_{n-1}, x_{n-2}) + (\gamma + \epsilon).$$
The condition on the roots ensures $\alpha + \beta < 1$.
Applying theorem's 2 and 3 from the reference we can bound solutions of the recursion $x_{n}$ between the following two recursions (taking maxes over chains of terms as shown in the article):
$u_{n} = (\alpha + \beta -\epsilon) u_{n-1} + (\gamma - \epsilon)$
and
$v_{n} = (\alpha + \beta +\epsilon) v_{n-1} + (\gamma + \epsilon)$
Since $\alpha + \beta < 1$ both {$u_{n}$} and {$v_{n}$} converge to the unique equilibrium.
Therefore given $\epsilon_{2} > 0$ there exists $M$ so that for $n>M$:
$$\frac{\gamma - \epsilon}{1-\alpha -\beta + \epsilon} - \epsilon_{2} \leq x_{n} \leq \frac{\gamma + \epsilon}{1-\alpha -\beta - \epsilon} + \epsilon_{2}$$
Thus $x_{n} \to \frac{\gamma}{1 - \alpha - \beta}$.
