A perturbation $\epsilon$ is by definition a very small "quantity". Therefore, the expansion of say, $(1+\epsilon)^n$ can be given by the well known Binomial series
$$(1+\epsilon)^n = 1+ n \epsilon + \frac{n(n-1)}{2} \epsilon^2 + \cdots $$
Similarly, if one has a $(\mathbf{1} + \textbf{T})^{-1}$ where $\mathbf{1}$ is an identity matrix and $\mathbf{T}$ such that $|\textbf{T}| < 1$ one can use the Neumann series $$(\mathbf{1} - \textbf{T})^{-1} = \sum_{k=0}^{\infty}\textbf{T}^k.$$
Question: Consider now the following expansion given in equation 12 of this article
$$\boxed{ \epsilon^{-1} \left[ \mathbf{1} - \frac{\textbf{T}}{\epsilon} \right]^{-1}= \epsilon^{-1} \sum_{k=0}^{\infty} \epsilon^{-k} \textbf{T}^{k} }$$ This expansion seems similar to the Neumann series. However, since $\epsilon^{-1} >>1$, how can this expansion converge? Is it at all a valid expansion?
First of all
$(\mathbf{1} + \textbf{T})^{-1} = \sum_{k=0}^{\infty}\textbf{T}^k$ is not correct.
Correct is $(\mathbf{1} - \textbf{T})^{-1} = \sum_{k=0}^{\infty}\textbf{T}^k$ if $|\textbf{T}| < 1.$
We have
$$ \epsilon^{-1} \left[ \mathbf{1} - \frac{\textbf{T}}{\epsilon} \right]^{-1}= \epsilon^{-1} \sum_{k=0}^{\infty} \epsilon^{-k} \textbf{T}^{k} $$
for $ \epsilon$ with $|\textbf{T}| < \epsilon.$