Matrix Inversion Test ( Sum of Matrix series) Friends,I have a set of matrices of dimension $3\times3$ called $A_i$.  ,
Following are the given conditions
a) each $A_i$ is non invertible except $A_0$ because their determinant is zero.
b) $\sum_{n=0}^\infty A_i$ is invertible and determinant is not zero
c) 


*

*This is the recursion available  for $A_i$,
$  A_{n}=\frac{1}{n} \{C_1* A_{n-1} +C_2 * A_{n-2}\} \tag 1$, where $A_0$ = Constant matrix ,$A_1$ =Constant matrix   

*$C_1,C_2 $ are constant matrices. $A_1$ and $A_0$ are  initial values. 
$A_0,A_1,C_1,C_2,A_n $  have dimension $3\times 3$

*$C_1,C_2,C_1+C_2 $ etc are skew symmetric matrices , not commutative, and also  with  diagonals as zeros  

*$A_n$   are converging series. Means last terms will be approaching to zero or very very small values

*Determinant of $C_1*A_{n-1}$ and $C_2*A_{n-2}$ both are zero {Logic : det($C_1A_{n-1}$)=det($C_1$)det($A_{n-1}$),=0*det($A_{n-1}$),$=0 $ }

*Given that SUM=  $  \sum_{n=0}^{n= \infty}   A_n \ne  0 $.

*Let $S(x) = \sum_{n=0}^\infty A_nx^n$,$SUM=S(1)$.Given that $S(1)$ is invertible  . Remember we still have not proved S(x) is invertible. What we only know is, S(1) is invertible  from the given conditions 
Question 
From the given condition can we say that $S(x)=\sum_{n=0}^\infty A_nx^n$ is invertible? If so. how do we prove that?. (x is not a matrix, it is just a variable)
 A: The matrix
$$
A(x):=\sum_{n=0}^\infty A_n x^n
$$
is invertible at $x=1$. The set of invertible matrices is an open set in the space of all matrices. Thus, an idea to solve the problem is to show that $x\mapsto A(x)$ is continuous. 
In order to prove that you need additional assumptions. One would be to assume that the power series
$$
\sum_{n=0}^\infty \|A_n\|\cdot x^n
$$
has convergence radius $r \ge1$. Here, $\|\cdot\|$ is a matrix norm. Then the series
$$
\sum_{n=0}^\infty A_n x^n
$$
would be convergent for all $x$ with $|x|<r$, hence $A(x)$ would be continuous on $(-r,+r) \cup \{1\}$ (assuming $x\in \mathbb R$).
Then you obtain that $A_n(x)$ is invertible for $x$ near $1$. 
A: Note: The question changed dramatically after this answer was written.
Since $\sum_n A_n$ exists, we see that $A_n \to 0$ and hence the radius of convergence of $\|A_n\|$ is at least one, so
the function $f(x) = \sum_n A_i x^n$ is real analytic  on $(-1,1)$.
Furthermore, Abel's theorem (applied to each component) shows that $\lim_{x \uparrow 1} f(x) = \sum_n A_n$.
Then continuity of $\det$ shows that $f(x)$ must be invertible in some interval
$(1-\epsilon,1]$ with $\epsilon>0$.
However, it is possible that $f$ is not defined for $x>1$.
First let $B_n=(-1)^n{1 \over n} I$ (each $B_n$ is invertible, but I will fix that in a moment). If $x>1$, we see that $\|B_n x^n\| \to \infty$, hence the sum does not converge. Now let $A_{3n-2}$ be the $(1,1)$ element of $B_n$, 
$A_{3n-1}$ be the $(2,2)$ element of $B_n$ and
$A_{3n}$ be the $(3,3)$ element of $B_n$. Each $A_n$ is singular, and the sum does not converge for $x>1$.
