Autocorrelation function $\rho(s)$ of AR(p), when s goes infinity Let $\{X_t\}_{t\in\mathbb{Z}}$ is the stacionary autoregressive process of degree p (AR(p)), and autocorrelation function of AR(p) is $$\rho(s)=\phi_1\rho(s-1)+\phi_2\rho(s-2)+\dots+\phi_p\rho(s-p), \text{ for $s=1,2,\dots, p$}.$$
I should show that when $s\rightarrow \infty$ then $\rho(s)\rightarrow0$. I am trying to solve difference equation and then $s\rightarrow \infty$. But I have no idea how to make it in general, because I think that when $s$ goes to the infinity ($s\rightarrow \infty$), it means also that p goes to the infinity ($p\rightarrow \infty$). Any help will be appreciated, thank you very much.
 A: This can be seen by solving for $\{\rho(s)\}_{s \geq 0}$ directly.
The system of linear homogeneous difference equation (of degree $p$) is
$$
\underbrace{ \rho(s)- \phi_1\rho(s-1)+\phi_2\rho(s-2)+\dots+\phi_p\rho(s-p) }_{ \Phi(L)\rho(s)} =  0, \;\; s \geq {0}.
$$
with initial conditions $\rho(0), \cdots \rho(p-1)$ given by the Yule-Walker equations
$$
\begin{bmatrix}
1           &\rho(1)      & \rho(2) &\cdots& \rho(p-1) \\
\rho(1)     & 1 & \rho(1) &\cdots   & \rho(p-2) \\
\rho(2)     & \rho(1)     & 1       &\cdots   & \rho(p-3) \\
\vdots      & \vdots      & \vdots  &\ddots   & \vdots \\
\rho(p-1)   & \rho(p-2)   &\rho(p-3)&\cdots   & 1 \\
\end{bmatrix}
\begin{bmatrix}
\phi_1           \\
\phi_2           \\
\phi_3           \\
\vdots           \\
\phi_p           \\
\end{bmatrix}
=
\begin{bmatrix}
\rho(1)           \\
\rho(2)           \\
\rho(3)           \\
\vdots           \\
\rho(p)           \\
\end{bmatrix}.
$$
The general solution is a linear combination of terms corresponding to the AR polynomial $\Phi(z)$.
For example, if $\Phi$ have $p$ distinct real roots $r_1, \cdots, r_{p}$,
the general solution is
$$
\rho(s) = c_1 r_1^{-s} + \cdots c_{p} r_{p}^{-s}.
$$ 
The coefficients $c_1, \cdots, c_{p}$ are given by the initial conditions.
By assumption (covariance stationarity), $|r_1|, \cdots, |r_{p+1}| > 1$, 
$$
\lim_{s \rightarrow \infty} r_l^{-s} = 0, \; l = 1, \cdots, p.
$$
This implies 
$$
\lim_{s \rightarrow \infty} \rho(s) = 0.
$$
The case of repeated or complex roots is similar. Same result holds. 
The key condition is that all roots of $\Phi$ lies outside the unit circle.
If there is a root on or inside the unit circle, e.g. if $r =1$ is a root, then the term $1^{-s}$ in $\rho(s)$ clearly does not approach $0$.
