If $T$ is an operator on $l_2$, and $\lambda>r(T)$ (where $r(T)$ denoted the spectral radius of $T$), the resolvent $(\lambda I-T)^{-1}$ can be expanded as $$ (\lambda I-T)^{-1}=\frac{1}{\lambda}I+\frac{1}{\lambda^2}T+\frac{1}{\lambda^3}T^2+\dots $$
If we fix some $x\in l_2$, and denote $y:=(\lambda I-T)^{-1}x$ the above expansion implies that $y$ is in the closed span of $(T^nx)_{n=0}^{\infty}$.
Is $y$ in this span even when $\lambda<r(T)$, so we do not have the Neumann series above to get that conclusion?
This is not true in general. Take $T$ to be the right shift operator on $l^2(\mathbb Z)$. It is invertible, but $T^{-1}x$ is not in the closure of the span of $(T^nx)_{n\ge0}$ for all $x\in l^2(\mathbb Z)\setminus \{0\}$ such that there is $K$ with $x_k=0$ for all $k<K$. (I.e. take $x=e_k$ to be a unit vector). I suspect that this holds for all $x\ne0$ but could not prove it.
If we assume that $\lambda$ belongs to the path-connected component of the resolvent set that contains $\{\mu:|\mu|>r(T)\}$ then the claim is true: $(\lambda I-T)x$ is in the closure of the span of $(T^nx)$ for $|\lambda|>r(T)$. If we have $\lambda_0$ in the resolvent set then for $\lambda$ close to $\lambda_0$, $(\lambda I-T)^{-1}$ can be written as power series in $(\lambda_0 I-T)^{-1}$. Then using path-wise connectedness and a covering argument, one should be able to prove the claim.