I have a question regarding this is from ProofWiki.
The number $b$ is known as the number base to which $n$ is represented. $n$ is thus described as being (written) in base $b$. Thus we can write $$ n = \sum_{j \mathop = 0}^m {r_j b^j} $$ as $ [r_m r_{m - 1} \ldots r_2 r_1 r_0 ]_b $
In the sum the lower bound is $0$ and the upper bound is $m$. The sum should be equal to $[r_m r_{m - 1} \ldots r_2 r_1 r_0 ]_b$, i.e. the first $r_j$ is $j=m$ (the upper bound). But in that case, shouldn't the bounds be reversed? So $$ n = \sum_{j \mathop = m}^0 {r_j b^j} $$ and $[r_m r_{m - 1} \ldots r_2 r_1 r_0 ]_b$?
Since addition on the integers is commutative, it doesn't matter in which order we add.
Either way the two sums you give do not always yield the same result. In the second sum $$\sum\limits_{j=m}^0 $$ we would do an addition and then increase $j$ by 1 as long as it is $\leq$ the number written on top of $\sum$. In our case that would mean that if $n\geq b$, we would stop already before doing the first addition, because $m>0$. By convention the sum would then evaluate to $0$.
You might want to read up on Capital sigma notation.