If A is invertible and $||B-A|| < ||A^{-1}||^{-1}$ prove $B$ is invertible. Just having really hard time trying to proof :
If $A$ is invertible and $||B-A|| < ||A^{-1}||^{-1}$ prove $B$ is invertible.
It is related to Neumann Series but i don't understand how to proof with math.
Thanks for your help and time.
Brian Ignacio

This is a general result in a Banach algebra. The algebra of linear operators on a finite dimensional space is a Banach algebra, but you also have many examples of infinite dimensional Banach algebra. For example the algebra of all bounded continuous real functions on some compact space (with pointwise operations and supremum norm) is a Banach algebra.
The norm on a Banach algebra satisfies the inequality $$\Vert A B \Vert \le \Vert A \Vert \Vert B \Vert.$$
Based on that, you can use the Neumann series, as suggested in your original question. For $\Vert U \Vert < 1$ and $n \in \mathbb N$ you have
$$(1-U)\sum_{k=0}^n U^k =U^{n+1} \tag{1}$$ and $\sum_{k=0}^n U^k$ converges normally to an element $$S=\sum_{k=0}^\infty U^k$$ and the equality (1) proves that $S$ is the inverse of $1-U$.
Coming back to your original question and using the initial answer of Patrick Da Silva, you get that $BA^{-1}=U$ is invertible. Hence $B$ is also invertible.
