I am studying for finals and am looking over previous problems that I missed significant points on. My instructor doesn't comment on why things are incorrect and he doesn't discuss homework, so I am on my own to figure out why I am wrong. Anyway, I would like some help figuring out where I went wrong.

Theorem:

If $\displaystyle A \text{ and } B \text{ are } n \times n$ matrices such that $\displaystyle \|I - AB\| < 1$, then $\displaystyle A \text{ and } $ are invertible. Furthermore, we have

$\displaystyle A^{-1} = B \sum_{k=0}^{\infty}(I - AB)^k \text{ and } B^{-1} = \sum_{k=0}^{\infty}(I - AB)^k A$

Problem:

Show that if $\displaystyle \|AB - I\| = \varepsilon < 1,$ then

$\displaystyle

\|A^{-1} - B \| \le \|B\|\left(\frac{\varepsilon}{1-\varepsilon}\right)

$

Proof:

By the theorem above we have

$\displaystyle \|A^{-1}\| = \|B \sum_{k=0}^{\infty}(I - AB)^k\| $

$\displaystyle \|A^{-1}\| \le \|B\| \sum_{k=0}^{\infty}\|I - AB\|^k $

$\displaystyle \|A^{-1}\| \le \|B\| \sum_{k=0}^{\infty}\varepsilon^k $

$\displaystyle \|A^{-1}\| \le \|B\| \frac{1}{1 - \varepsilon} $

$\displaystyle \|A^{-1}\| + \|-B\| \le\|B\| \frac{1}{1 - \varepsilon} + \|B\|$

$\displaystyle \|A^{-1} - B\| \le \|B\|\frac{2 - \varepsilon}{1 - \varepsilon}$

Which doesn't seem to get me anywhere. Any help is appreciated. Thank you in advance.