Exercise 5.17 (Hunter and Nachtergaele) Suppose that $K: X \rightarrow X$ is a bounded linear operator on a Banach space $X$ with $\|K\|<1 .$ Prove that $I-K$ is invertible and $$ (I-K)^{-1}=I+K+K^{2}+K^{3}+\ldots $$ where the series on the right hand side converges uniformly in $\mathcal{B}(X)$.
My approach: I think we should use open mapping theorem in some form because it talks about inverses? But then we need to show the inverse exists instead of saying something about boundedness of the inverse, which is what open mapping theorem does, so I'm a bit confused.
Just verify that the series $\sum_n K^{n}x$ converges for each $x$ using completeness of $X$ and the fact that $\|K^{n}x\| \leq \|K\|^{n} \|x\|$. Then define $Tx =\sum_n K^{n}x$. Uniform Boundeness Principle gurantees that $T$ is a bounded operator: Apply the theorem to the sequence of operators $(T_n)$ where $T_n (x)= \sum\limits_{i=1}^{n} K^{i}x$. Now it is an easy matter to verify that $(I-K)T=T(I-K)=I$ so $T$ is indeed the inverse of $I-K$.