$\lVert A\lVert<1$ implies $1-A$ invertible only true in complete spaces? It is a well known fact that 

if in a Banach space $X$ a bounded linear operator $A:X\to X$ satisfies $\lVert A\lVert<1$, then $1-A$ has a bounded inverse.

I was wondering wether completeness is actually a necessary condition. I think the following argument is corect and doesn't use completeness:
Clearly for any $N\in\mathbb{N}$ $$\left(\sum_{k=0}^N A^k\right)(1-A)=1-A^{N+1}.$$ 
Since $\lVert A^{N+1}-0\lVert\leq\lVert A\lVert^{N+1}\rightarrow 0$ the limit of the right hand side exists and equals $1$. Hence also $$\lim_{N\rightarrow\infty}\left(\sum_{k=0}^N A^k\right)(1-A)=1$$ and therefore $(1-A)^{-1}=\lim_{N\rightarrow\infty} \sum_{k=0}^N A^k$. To see that the inverse is bounded I note that by continuity of the norm $$\lVert(1-A)^{-1}\lVert=\lim_{N\rightarrow\infty} \left\lVert \sum_{k=0}^NA^k\right\lVert\leq\sum_{k=0}^N \lVert{A}\lVert^k=\frac{1}{1-\lVert A\lVert}<\infty.$$ 

Without some completeness hypothesis, the infinite sum expressing the inverse may not converge in the given space.
