<s>[INST] Why do I get $0.098765432098765432...$ when I divide $8$ by $81$? I got this remarkable thing when I divided $16$ by $162$, or, in a simplified version, $8$ by $81$.  It's $0.098765432098765432\cdots$, or more commonly known as $0.\overline{098765432}$, with all the one-digit numbers going backwards...except for $1$.  Yeah, it's missing the $1$.  One, how do I get this remarkable outcome and two, why is it missing the $1$?
 [/INST] More generally, in base $b$ (any integer $>1$)
$$ \dfrac{b-1}{b^2} + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} = 
\dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}$$
so that
$$ \eqalign{
\dfrac{b-1}{b^2}& + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} +
\dfrac{0}{b^{b}} + \dfrac{b-1}{b^{b+1}} + \dfrac{b-2}{b^{b+3}} + \ldots
= \sum_{j=0}^\infty b^{-j(b-1)} \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}\cr
&= \dfrac{b^{b-1}}{b^{b-1}-1} \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}= \dfrac{(b-2)}{(b-1)^2}}$$
What you have is the case $b=10$.
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