I have troubles understanding a step in a proof.
Let $\pm\sum_{n=-k}^{\infty}a_nb^{-n}$ be a b-adic number with $a_n,b,k\in\mathbb{N}$, $b\ge 2$ and $0\leq a_n<b$. We want to prove that this number is a Cauchy-Seqence. The prove given in my textbook now uses the partial-sums $x_n:=\sum_{v=-k}^{n}a_vb^{-v}$ and shows, that (x_n) is a Cauchy-Sequence. Given $\epsilon>0$ and $N\in\mathbb{N}$ so that $b^{-N}<\epsilon$. Then we have for $n\ge m\ge N$ that
$$|x_n-x_m| =\sum_{v=m+1}^{n}a_vb^{-v}\leq \sum_{v=m+1}^{n}(b-1)b^{-v}$$
The proof continues but my problem is with the last step in the above equation.
I know and tried that $a_vb^{-v}\leq bb^{-v}$ for all $a_v$, $(b-1)b^{-v}=b^{1-v}-b^{-v}$ and $\sum_{v=m+1}^{n}(b-1)b^{-v}=b^{-m}-b^{-n}$. But all these don't let me understand this inequality. Can anyone explain this to me? Thanks in advance. Please let me know if additional context is needed.
Since $a_n,b\in\mathbb{N}$, the given condition $a_n<b$ implies $$ a_n\le b-1$$ for all $n$.
In other words, the greatest integer that is less than an integer $b$ is $b-1$.