Let $X $ be a Banach space $(X,\|.\|)$ with $\dim(X)=\infty $ and $T\in B(X) $.
Suppose that $\left\| I-T\right\| < 1$. Show that $T^{-1}\in B\left( X\right)$ and $\left\| I-T^{-1}\right\| \leq \dfrac {\left\| I-T\right\| } {1-\left\| I-T\right\| }$.
The condition $\|I-T\|<1$ allows you to define $$S=\sum_{k=0}^\infty (I-T)^k. $$ Now check that $(I-T)S=S-I $. This is exactly $TS=I $. And since $S $ commutes with $T $, you get that $T $ is invertible and $T^{-1}=S $.
The inequality is nothing but the fact that $0 <\|I-T\|<1$. If you add one to the left inequality you get $1 <1+\|I-T\|$, so $1-\|I-T\|<1$. Since $1-\|I-T\|>0$, we can reverse the inequality (or divide, whatever you prefer) to get $$1 <\frac1 {1-\|I-T\|}. $$ Now multiply both sides by $\|I-T\|$.