Show that $(A+B)^{-1} -A^{-1} = A^{-1}\sum_{k=1}^{\infty}({BA^{-1}})^{k}$ goes to zero as $B$ goes to zero.
This leads rise to another question of mine, can I say that, if $||BA^{-1}|| \leq r < 1$, then the series: $A^{-1}\sum_{k=1}^{\infty}(-1)^{k}({BA^{-1}})^{k}$ converges at a rate dependent only on r.
My attempt:
Well, we can say if $||BA^{-1}||$ is less than 1, then this converges with the geometric series $A^{-1}\sum_{k=1}^{\infty}||({BA^{-1}})^{k}||$.
$(A+B)^{-1} -A^{-1} \leq ||A^{-1}\frac{BA^{-1}}{1-BA^{-1}}||$ which can we say goes to zero as B goes to zero.