Power series expansion of perturbed matrix inverse Is there a power series expansion for a matrix inverse of the form $\left(A+D\right)^{-1}$, where $D$ is a small perturbation of A? $A$ is invertible. 
 A: Yes.  
You can wtite
$A + D = (I + DA^{-1})A; \tag 1$
then if $\Vert D \Vert$ is sufficiently small that
$\Vert DA^{-1} \Vert \le \Vert D \Vert \Vert A^{-1} \Vert < 1, \tag 2$
which occurs if
$\Vert D \Vert < \Vert A^{-1} \Vert^{-1}, \tag 3$
we have the series representation 
$(I + DA^{-1})^{-1} = \displaystyle \sum_0^\infty (-DA^{-1})^k; \tag 4$
this series converges by vrtue of (2), being majorized by
$ \displaystyle \sum_0^\infty \Vert DA^{-1} \Vert^k = \dfrac{1}{1 - \Vert DA^{-1} \Vert} < \infty, \tag 5$
the equality here being the standard formula for the sum of an infinite geometric series.
From (1),
$(A + D)^{-1} = ((I + DA^{-1})A)^{-1} = A^{-1}(I + DA^{-1})^{-1}; \tag 6$ 
substituting (4) on the right-hand side yields
$(A + D)^{-1} = ((I + DA^{-1})A)^{-1} = A^{-1}\displaystyle \sum_0^\infty (-DA^{-1})^k =\displaystyle \sum_0^\infty A^{-1} (-DA^{-1})^k,    \tag 7$ 
the desired series for $(A + D)^{-1}$.
Note Added in Edit, Friday 13 December 2019 8:11 AM PST: It is probably worth mentioning how the series (4) may be validated.  Let $X$ be any operator such that
$\Vert X \Vert < 1; \tag 8$
then basic algebra shows us that
$(I - X) \displaystyle \sum_0^n X^i = (I - X)(I + X + X^2 + \ldots + X^n) = I - X^{n + 1}; \tag 9$
by virtue of (8),
$\Vert X^{n + 1} \Vert \le \Vert X \Vert^{n + 1} \to 0 \; \text{as} \; n \to \infty; \tag{10}$
therefore (9) yields, again with $n \to \infty$,
$(I - X) \displaystyle \sum_0^\infty X^i = I, \tag{11}$
whence
$(I - X)^{-1} = \displaystyle \sum_0^\infty X^i; \tag{12}$
taking 
$X = -DA^{-1} \tag{13}$
in (12) yields (4).  End of Note.
A: Let $A$ be nonsingular and let $\tilde{A}=A+E$ be a perturbation of $A$. It can be shown that 
\begin{align}
    \tilde{A}^{-1}={A}^{-1}-{A}^{-1}E\tilde{A}^{-1} \tag 1 
\end{align}
The main drawback of this formula is the presence of $\tilde{A}^{-1}$ on both sides of the equality sign. However, by expressing $\tilde{A}^{-1}$ into infinite series, it can be written 
\begin{align} \label{taylorExpansion_of_A}
    \tilde{A}^{-1}= {A}^{-1}-({A}^{-1}E){A}^{-1}+({A}^{-1}E)^2{A}^{-1}-\cdots \tag 2
\end{align}
Since, ${A}^{-1}$ is a differential function of its elements, first order approximation of (\ref{taylorExpansion_of_A}) is accurate up to terms of order $||E||^2$. With this approximation (\ref{taylorExpansion_of_A}) can be written as, 
\begin{align}\label {tag 3}
    \tilde{A}^{-1}= {A}^{-1}-({A}^{-1}E){A}^{-1}. \tag 3
\end{align}
It appears that the answer give by @Robert Lewis is similar, where (\ref{tag 3}) is truncaded. 
