Exceptions to special case of binomial inverse theorem involving $(A-B)^{-1}$?

Question

Under the Woodbury matrix identity page on Wikipedia, there is a special case under the binomial inverse section where $$(A-B)^{-1}=\sum_{k=0}^\infty (A^{-1}B)^{k}A^{-1}\label{a}\tag{1}$$

In my case, I was trying to apply it where $A=I$, which I believe simplifies nicely to $$=\sum_{k=0}^\infty B^k\label{b}\tag{2}$$

At first, it seemed to work fine, but when testing it with random data sets, I started get obscenely large numbers in the resulting matrix. A simple case of when it seems to fail consistently is when the values of the square matrix $B$ are in the interval $[1,\infty)$. At that point, the simplification seems to lead all elements of the result to infinity as $k$ goes to infinity, which when summed together, is clearly is not the inverse.

I am able to verify this in code (using R) just generating random values for $B$. I can solve for the inverse of $(I-Q)$ using built in functions and confirm the results by multiplying it by $(I-Q)$ to get $I$, but when I try to manually calculate using $\ref{b}$ it falls apart.

Are there exceptions or rules somewhere for $\ref{a}$ that I'm missing and inadvertently violating?

You have a similar problem even with the binomial theorem over real numbers; take $(1 - 2)^{-1},$ for example, and expand it by the binomial formula. — David K, Dec 20 '18 at 16:15
The infinite sum $\sum_{k=0}^\infty C^k$ converges if and only if $|C|<1$, so $|A^{-1}B|<1$ is a necessary condition you need for (1) to hold. — Mike Earnest, Dec 20 '18 at 16:24
@MikeEarnest Norm bound by one is sufficient, but not necessary for convergence. — A.Γ., Dec 20 '18 at 16:29

score 4 · Answer 1 · answered Dec 20 '18 at 16:31

We will indeed have $(I - B)^{-1} = \sum_{k=0}^\infty B^k$ whenever the sum on the right converges. This sum will converge if and only if the eigenvalues of $B$ all have magnitude strictly less than $1$.

The equation on wikipedia should have a blurb to the effect of "whenever the sum on the right converges".

A $1 \times 1$ example that illustrates what's going on: note that $$ \frac 1{1-x} = 1 + x \cdot \frac{1}{1 - x} $$ this equation has a "recursive structure", so we could also write $$ \frac 1{1-x} = 1 + x \cdot \left[1 + x \cdot \frac{1}{1 - x}\right]\\ = 1 + x + x^2 \cdot \frac 1{1-x} $$ so that in general, we have $$ \frac 1{1 - x} = 1 + x + x^2 + x^3 + \cdots + x^n \cdot \frac 1{1-x} $$ Now, whenever the expression on the right approaches a limit, we could also write $$ \frac 1{1 - x} = 1 + x + x^2 + x^3 + \cdots = \sum_{k=0}^\infty x^k $$ this manipulation only makes sense rigorously (i.e. says something accurate in terms of the usual notion of a limit) when $|x| < 1$.

Exceptions to special case of binomial inverse theorem involving $(A-B)^{-1}$?

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