Proving an operator from $\ell^\infty \to \ell^\infty$ is invertible (Just to fix some notation, $\ell^\infty = \ell^\infty(\mathbb{N\cup \{0\}},\mathbb{C})$ in what follows)
Let $S:\ell^\infty\to\ell^\infty$ be the left shift operator and let $A:\ell^\infty\to\ell^\infty$ be the the multiplication by the element $a\in \ell^\infty$. 
In other words we have that $S(x)_n = x_{n+1}$ and $A(x)_n=a_n x_n$.
Suppose that $a\in \ell^\infty$ satisfies
$$(1)\quad \quad |\Pi_{k=0}^{n-1}a_{j+k}|\leq c\alpha^n \ \ \forall j\geq 0$$ 
where $c>0, \alpha \in (0,1)$.
My problem is the following:

Prove that $I-AS$ is invertible ($I$ is the identity operator).

Unfortunately $\|AS\|_{\mathcal{\ell^\infty}}= \|a\|_{\ell^\infty}$ that can be greater than $1$ so we cannot exploit the Neumann series to define the inverse.
I managed to prove, though, that $I-AS$ is injective, indeed if $y = AS y$ then we would have (by $(1)$) that 
$$y_n = a_n y_{n+1} = a_n a_{n+1}\dots a_{n+N} y_{n+N+1}\leq ||y|||_{\infty}\Pi_{k=0}^{N}a_{n+k}|\leq ||y|||_{\infty} c\alpha^{N+1} $$
for any $N>0$ thus taking the limit $y_n = 0 $ for any $n$.
Now this would be sufficient to conclude if for example we manage to prove that $AS$ is a compact operator. But this is not the case since the image of the unit ball through $S$ is the ball itself and $A$ is not compact (for example consider a sequence $a$ such that $a_{2n} = 0$ $a_{2n+1}=1$ so that condition $(1)$ is satisfied).
The only way I see to solve this problem is therefore to prove that $I-AS$ is surjective or maybe that is Fredholm of index $0$.
And here I got stuck since I tried to write the inverse  obtaining
$$ (I-AS)^{-1}(y)_n = y_n + \sum_{k\geq 0} a_{n+k} y_{n+k+1}$$ 
but I don't see a way to exploit relation $(1)$ to prove the series is convergent.
 A: It seems that there is a problem in your computation of the inverse. Indeed, let $\left(y_n\right)_{n\geqslant 0}$ be an element of $\ell^\infty$. Then we would like to prove that there exists some $x\in\ell^\infty   $ such that $x-ASx=y$, which is equivalent to 
$$x_n=a_nx_{n+1}+y_n \mbox{ for each }n\geqslant 0.       $$
One can prove by induction that for any $N\geqslant 1$, 
$$\tag{* }   x_n=y_n+\sum_{i=1}^Ny_{n+i}  \prod_{l=0}^{i-1}a_{n+ l}   +x_{n+N+1}\prod_{l=0}^Na_{n+l}. $$
Using the assumption (1) in the opening post and the fact that $\left\lvert y_{n+i}\right\rvert \leqslant \left\lVert y\right\rVert_\infty$ for each $n$ and $i$, one get the convergence of  $\sum_{i=1}^{+\infty}   y_{n+i}  \prod_{l=0}^{i-1}a_{n+ l}$. Since $\left\lvert x_{n+i}\right\rvert$ is bounded independently of $n$ and $i$, an application of (1) gives that $x_{n+N+1}\prod_{l=0}^Na_{n+l}$ goes to $0$ as $N$ goes to infinity. Therefore, letting $N$ going to infinity in (1), one obtains that 
$$x_n=y_n+\sum_{i=1}^{+\infty}   y_{n+i}  \prod_{l=0}^{i-1}a_{n+ l}.$$
Using again (1), one gets that $x_n$ is bounded independently of $n$.     
