Let $E$ be a Banach space and $A\in L(E)$ such that $\parallel A \parallel<1$. Then, $I-A$ is invertible and $(I-A)^{-1}=\sum^\infty_{n=1}A^n$. I want to prove the following:

Let $E$ be a Banach space and $A\in L(E)$ such that $\parallel A \parallel<1$. Then, $I-A$ is invertible and $(I-A)^{-1}=\sum^\infty_{n=1}A^n$.

I already proved that the series $\sum^\infty_{n=1}A^n$ converges. Now, I just have to prove that $\sum^\infty_{n=1}A^n(I-A)=I$.
I have $\sum^\infty_{n=1}A^n(I-A)=\lim_N\rightarrow\infty \sum^N_{n=1}(A^n-A^{n+1})$, so I have to check that this limit is $I$.
Let $\epsilon>0$. I have $$\parallel \sum^N_{n=1}(A^n-A^{n+1})-I \parallel \le \sum^N_{n=1}\parallel A^n \parallel + \sum^N_{n=1}\parallel A^{n+1} \parallel +\parallel I \parallel \\ \le \sum^N_{n=1}(\parallel A\parallel^n  + \parallel A\parallel^{n+1}) +1  $$
But now large $N$ does not help me getting arbitrarily small values (less than $\epsilon$)... quite the contrary.
What to do?

Taking your notations, we already have $\sum A^n$ that converges in $L(E)$ to some element $B$ (also, note the comment that the series should begin at $n=0$).
Now, we also have the following telescopic sum:
$$
\left(\sum_{n=0}^N A^n\right) \circ (I - A) = I - A^{N+1}
$$
The left hand side converges to $B \circ (I-A)$, the right hand side converges to $I$ (because $||A^{N+1}||\leq||A||^{N+1} \longrightarrow 0$ as $N \rightarrow \infty$).
Uniqueness of the limit tells us that $B \circ (I - A) = I$. The same process can be done the other way around to show that $(I-A) \circ B = I$. 
