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).