 Solution

The bounded operators on a Banach space form a Banach algebra in the operator norm. If $\Vert T\Vert <1$, the Neumann series $A={\sum }_{k=0}^{\infty }{T}^k$ converges absolutely in that norm. Its partial sums satisfyTaking limits proves $\boxed{(I-T{)}^{-1}=\sum _{k\ge 0}{T}^k}$.

If $|\lambda |>\Vert T\Vert$, factor $\lambda I-T=\lambda (I-T/\lambda )$ and apply the same argument. Thus such $\lambda$ are in the resolvent set and