Let $H$ be an Hilbert space, $A:H\to H$ be a bounded linear operator such that $$ \|A^{n_0}\|< 1\qquad\text{for some}\quad\; n_0\in\mathbb{N}. $$ I have to show that $I-A$ is invertible.
My idea is to consider $$ S:=\sum_{n=0}^\infty A^n, $$ where $A^n$ is defined by induction as usual.
Then $$ (I-A)S=\lim_{n\to\infty} I-A^{n+1}. $$
We have just to show $$ \lim_{n\to\infty}A^{n+1}=0. $$
Any ideas?
Hint: \begin{align} S = \sum_{m=0}^{\infty} \sum_{\ell = 0}^{n_{0} -1} A^{mn_{0}+\ell} \end{align}