If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm < $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$.
Thinking in terms of a converse, if $T$ is any bounded linear operator defined on $X$, then does the existence of a bounded inverse $S=(I-T)^{-1}$ imply that $S$ can be represented as $S=\sum_{n=0}^\infty T^n$?
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Well there are slightly weaker condition that improves the result you cite.
Suppose $T$ is a bounded operator with spectral radius $r(T)< 1$ Then by the spectral radius formula we have $$\lim_{n\to\infty}\|T^n\|^{1/n} =\inf_n\|T^n\|^{1/n}=r(T)< 1$$ which ensure the convergence of $$\sum_{n=0}^\infty\|T^n\|$$ which in turn, by the triangle inequality, bounds $$\left\|\sum_{n=0}^\infty T^n\right\|$$ Now, it is a standard exercise to show that $$\sum_{n=0}^\infty T^n=(I-T)^{-1}$$ If there is any doubt at all do not hesitate to ask.. **Edit:** By the Banach algebra inequality we have $\|T^n\|\le\|T\|^n$, which means that $$r(T)=\inf_k\|T^k\|^{1/k}\le\|T^n\|^{1/n}\le\|T\|$$ Hence $\|T\|<1$ implies not only $r(T)<1$, but also $r(T)\le\|T\|<1$. Also, this is not the case in the example of Robin above, because we also have $r(T)=\sup{|\lambda|:\lambda\in\sigma(T)}$ where $\sigma(T)$ is the spectrum of $T$ (the set of all $\lambda\in\mathbb{C}$ such that $\lambda I-T$ is not invertible) and the eigenvaules of $T$ is certainly in the spectrum. (Note that $r(T)$ is the radius of the smallest closed disc that contain $\sigma(T)$ - hence the name spectral radius).
No, not even in the finite-dimensional case. If $T$ is a linear map from $\mathbb{R}^n$ to itself with all eigenvalues $>1$ in absolute value, then $I-T$ is invertible, but $\sum T^n$ certainly does not converge.