Finding the limit of a Matrices determinant The problem is as follows:

I've been trying to figure this out with no luck. I'm lost at the $A_k+1$ and $A_0$. I'm not sure what they are implying and how they would apply in finding the limit. 

Note that it can be easily shown that 
$$
A^{-1} \;\; =\;\; \left [ \begin{array}{cc}
2 & -1\\
-1 & 1 \\
\end{array} \right ].
$$
This implies that 
$$
A_1 \;\; =\;\; \left [ \begin{array}{cc}
3/2 & 0 \\
0 & 3/2 \\
\end{array} \right ]
$$
Implying that 
$$
A_2 \;\; =\;\; \left [ \begin{array}{cc}
3/4 & 0 \\
0 & 3/4 \\
\end{array} \right ] + \left [ \begin{array}{cc}
1/3 & 0 \\
0 & 1/3 \\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
13/12 & 0 \\
0 & 13/12 \\
\end{array} \right ]
$$
Implying that 
$$
A_3 \;\; =\;\; \left [ \begin{array}{cc}
13/24 & 0 \\
0 & 13/24 \\
\end{array} \right ] + \left [ \begin{array}{cc}
6/13 & 0 \\
0 & 6/13 \\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
313/312 & 0 \\
0 & 313/312 \\
\end{array} \right ].
$$
Therefore we have $A_3^{-1} = \frac{312}{313}I$, and thus $\det(A_3^{-1})^n = \left (\frac{312}{313}\right )^{2n}$ which tends to zero as $n\to \infty$.  
