A criterion for invertibility of a bounded linear operator. I'm studying Semigroup Theory and I wasn't able to understand a step in a certain proof. As far as I have been able to understand, the author used the following result:
If $A$ is a bounded linear operator and $\|A-I\|<1,$ then $A$ is invertible.
Is it true? If it is true, how can I prove it.
 A: It is true, provable as follows:  for any operator $B$ with $\Vert B \Vert < 1$, $I - B$ is invertable since
$(I - B)(\sum_0^n B^k) = I - B^{n + 1}; \tag{1}$
now note that $\Vert B \Vert < 1$ implies
$\Vert B^n \Vert \le \Vert B \Vert^n \to 0 \tag{2}$
as $n \to \infty$;  thus 
$B^n \to 0 \tag{3}$
as $n \to \infty$ as well.  Letting $n$ increase without bound in (1) thus yields
$(I - B)(\sum_0^\infty B^k) = I; \tag{4}$
that is, $I - B$ is invertible with inverse $\sum_0^\infty B^k$.
Taking $B  = I - A$ establishes the result.  
Note Added Sunday 10 May 2015 11:10 AM PST:  It's probably not a bad idea, for the sake of completeness, to flesh things out just a tad and show explicitly that $\sum_0^\infty B^k$ converges, although I think this might follow from the above without further words on the matter.  Nevertheless, setting
$S_n = \sum_0^n B^k \tag{5}$
we have for $m > n$
$\Vert S_m - S_n \Vert = \Vert \sum_{n + 1}^m B^k \Vert =  \Vert B^{n + 1}\sum_0^{m - n - 1} B^k \Vert \le \Vert B \Vert^{n + 1} \Vert \sum_0^{m - n - 1} B^k \Vert$
$\le  \Vert B \Vert^{n + 1}(\sum_0^{m - n -1} \Vert B \Vert^k) \le \Vert B \Vert^{n + 1} (\sum_0^\infty \Vert B \Vert^k) = \dfrac{\Vert B \Vert^{n + 1}}{1 - \Vert B \Vert}, \tag{6}$
where $\sum_0^\infty \Vert B \Vert^k$ converges, and is equal to $(1 - \Vert B \Vert)^{-1}$, since $\Vert B \Vert < 1$; it is a real geometric series with ratio $\Vert B \Vert$.  Lettting $n \to \infty$ then shows that $S_n$ is Cauchy etc. etc. etc.  End of Note.
A: This comes from the fact that if $|x|<1$ then $1+x+x^2+\cdots = {1 \over 1-x}$.
Let $y=x-1$ to get
$1+(1-y)+(1-y)^2+\cdots = {1 \over y}$.
Write $A=I +(A-I)$, and look at the series $I+(A-I)+(A-I)^2+\cdots$. Show that this converges and the result equals $A^{-1}$.
