Comparing asymptotics in the constant/nonconstant case for linear recurrence of order 2 Let $(u_n)_{n\geq 1}$ be a sequence of real numbers satisfying a (non-constant) linear recurrence of order $2$ :
$$
u_{n+2}+a_nu_{n+1}+b_nu_n=0 \tag{1}
$$
where $(a_n)$ and $(b_n)$ are both convergent sequences, with limit $a$ and $b$ respectively. Suppose further that any solution of the constant recurrence tends to zero (in other words, the two roots of $X^2+aX+b$ have modulus $<1$).
Question 1. Must $(u_n)$ converge to zero as in the constant case ?
Question 2. (independent of question 1) When $(u_n)$ converges to zero, must we have a geometric convergence $|u_n| \leq kc^n$ with $0 \lt c \lt 1$ as in the constant case ?
Update 12/09/2019 : The answer to both questions is yes when $|a|+|b| \lt 1$, because then there is a $r\in (0,1)$ such that $|a_n|+|b_n| \leq r$ for large enough $n$, say $n\geq n_0$ for some $n_0\in{\mathbb N}$. Then, (1) yields $|u_{n+2}| \leq r \times {\mathsf{max}}(|u_n|,|u_{n+1}|)$ for all $n\geq n_0$. Putting $k={\mathsf{max}}(|u_{n_0},u_{n_0+1}|)$, by induction we deduce that $|u_{n_0+p}| \leq kr^{\lfloor \frac{p}{2} \rfloor}$ for all $p\geq 0$.
 A: The answer is yes for both questions.
The scalar recursion of order 2 is equivalent to the following recursion
of order 1 for vectors via $v_n=(u_n,u_{n+1})^T$.
\begin{equation}\tag{1}\label{one}
v_{n+1}=A_nv_n,\ \ A_n=\left(\begin{matrix}0&1\\-b_n&-a_n\end{matrix}\right)
\end{equation}
If $\lambda,\mu$ are the roots of $X^2+aX+b$ which satisfy $|\lambda|,|\mu|<1$ by assumption  then there is a constant matrix $T$ such that 
$$\newcommand{\eps}{\varepsilon}T^{-1}A T= \left(\begin{matrix}\lambda&\eps\\0&\mu\end{matrix}\right)=:D
\mbox{ for }A=\lim A_n=\left(\begin{matrix}0&1\\-b&-a\end{matrix}\right)$$
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
$$w_{n+1}=(D+F_n)w_n,$$
where $F_n$ is a matrix of elements tending to 0.
Using the maximum norm for vectors and the corresponding matrix norm, we find
$$||w_{n+1}||\leq\big(\max(|\lambda|+|\eps|,|\mu|)+||F_n||\big)||w_n||.$$
By assumption, the maximum is smaller than 1 and $||F_n||$ tends to 0. Therefore there
exist some $0<M<1$ and some integer $N$ such that the big parenthesis is $\leq M$ for $n\geq N$. This means that
$$||w_{n+1}||\leq M||w_{n}||\mbox{ for }n\geq N$$
and hence $||w_n||\leq ||w_N|| M^{n-N}$ for $n\geq N$.
Therefore $w_n$ tends to 0 and therefore also $v_n$ and $u_n$ tend to 0 and there
are constants $K,M$, $M<1$ such that $|u_n|\leq K M^n$ for  all $n$.
