# Limit of the determinant of a series of matrices

Given the $N\times N$ matrix $A$, consider the series: $$B=\sum_{k=1}^{N}(A^k)^{-1}$$ where the symbol $o^{-1}$ means the inverse of $A^k$ is it possible and if yes how, to find all the matrices for which the quantity $$C=\lim_{N\to\infty}(\det(B))$$ has a finite value? We can consider for the sake of simplicity the case $N=2$. Thanks.

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I guess the number of terms in the sum should be independent of the dimension of the matrix. – Davide Giraudo Sep 25 '12 at 12:20
The formula for geometric series should be of use here. – Marc van Leeuwen Sep 25 '12 at 12:44

If $\det(A-\lambda\cdot I)\ne0$ for every $\lambda\in\mathbb C$ such that $|\lambda|\leqslant1$, then $$C=\frac1{\det(A-I)}$$
To show this, assume that $B_n\to S$ and note that $C=\det(B_n)\to\det(S)$ and that $S$ should satisfy $$AS=A\sum\limits_{k=1}^{+\infty}A^{-k}=I+S,$$ hence $\det(S)\cdot\det(A-I)=\det(I)=1$. For this convergence to happen, one should assume that $A$ has no eigenvalue $\lambda$ such that $|\lambda|\leqslant1$.