Inverse of the Toeplitz matrix

By Cramer's rule, $U_t=A_t^{-1}(1,1)$ is given by the ratios of two determinants. Let:

$$A_3=\left(\begin{array}{ccc}c+1&\rho&\rho^2 \\ \rho&c+1&\rho\\ \rho^2&\rho&c+1\end{array}\right),\qquad B_3=\left(\begin{array}{ccc}1&\rho&\rho^2 \\ \rho&c+1&\rho\\ \rho^2&\rho&c+1\end{array}\right)$$ and define $A_t,B_t$ in a similar fashion. By replacing the first column with the first column minus $\rho$ times the second column, then expanding along the first column, we get: $$ \det A_{t+1} = (c+1-\rho^2) \det A_t + c\rho^2 \det B_t, $$ $$ \det B_{t+1} = (1-\rho^2) \det A_t + c\rho^2 \det B_t, $$ and by eliminating $B_t$, $$ \det A_{t+1} = (c+1+c\rho^2-\rho^2) \det A_t -c^2\rho^2\det A_{t-1} $$ then dividing the whole line by $\det A_t$, $$ \frac{1}{U_{t+1}} = (c+1+c\rho^2-\rho^2)-c^2\rho^2 U_{t} $$ or:

$$ U_{t+1} =\frac{1}{(c+1+c\rho^2-\rho^2)-(c^2\rho^2)U_t}$$

where obviously $U_1=\frac{1}{c+1}$. If we set $U_t=\frac{V_t}{c\rho},$

$$ V_{t+1} =\frac{1}{\frac{c+1+c\rho^2-\rho^2}{c\rho}-V_t},$$ so by setting $\beta=\frac{c+1+c\rho^2-\rho^2}{c\rho}$ and assuming $\beta \geq 2$ we have: $$ \lim_{t\to +\infty}U_t = \frac{-\beta+\sqrt{\beta^2-4}}{2c\rho}$$ by the usual convergence properties of the negative continued fractions.