Show that $B_j$ is a Cauchy sequence with respect to the operator norm

Question

Given $A\in L(\mathbb R^n)$ such that $||A||<1$, set $B_j=\sum\limits_{r=0}^n A^r$.

Show that $B_j$ is a Cauchy sequence with respect to the operator norm.

$||B_m-B_n||=||\sum\limits_{r=0}^m A^r-\sum\limits_{r=0}^n A^r||=||\sum\limits_{r=n+1}^m A^r||\leq\sum\limits_{r=n+1}^m ||A^r||<m-n$

Not sure where to go from here, is this right so far?

EDIT:

$\sum\limits_{r=n+1}^m ||A^r||\leq\sum\limits_{r=n+1}^m ||A||^r=\sum\limits_{r=0}^m ||A||^r-\sum\limits_{r=0}^n ||A||^r=\frac{||A||^{n+1}-||A||^{m+1}}{1-||A||}$

Still can't see where this is going.

Answer 1

Write $m = n + p$ with $p \geq 1$. Then your right-hand side becomes $$ ||A||^{n+1} \frac{1 - ||A||^{p}}{1 - ||A||} $$ and $$ 0 \leq \frac{1 - ||A||^{p}}{1 - ||A||} \leq \frac{1}{1 - ||A||} $$ Now set $n$ large enough to bound the right-hand side with a variable $\varepsilon$.