There's quite a lot about this on the Internet.
After a gap of nearly 12 years, in which there seems to be nothing new, it emerged again on Twitter yesterday:
Quite Interesting on X: "If you divide 1 by 998001, the resulting decimal will feature every three-digit combination from 000 to 999 with the exception of 998."/ X.
There are a few other questions about it on Maths.SE:
here,
here,
here,
here.
But as far as I can tell, no-one has actually written out a proof of the infinite decimal expansion of $1/998001.$
Here is my attempt.
For integral $n\geqslant2,$ and real $x\ne1,$
\begin{align*}
1+2x+3x^2+\cdots+(n-1)x^{n-2}&=\frac{d}{dx}(1+x+x^2+\cdots+x^{n-1})
\\&=\frac{d}{dx}\frac{1-x^n}{1-x} \\
&= \frac{1-x^n}{(1-x)^2} - \frac{nx^{n-1}}{1-x} \\
&= \frac{1-nx^{n-1}+(n-1)x^n}{(1-x)^2}.
\end{align*}
In particular,
$$
1+\frac2n+\frac3{n^2}+\cdots+\frac{n-1}{n^{n-2}} =
\frac{1-n^{-(n-2)}+n^{-(n-1)}-n^{-n}}{(1-n^{-1})^2}.
$$
Therefore,
\begin{align*}
\frac1{n^2}+\frac2{n^3}+\frac3{n^4}+\cdots+\frac{n-1}{n^n} &=
\frac{1-n^{-(n-2)}+n^{-(n-1)}-n^{-n}}{(n-1)^2} \\
&=\frac{n^n-n^2+n-1}{(n-1)^2n^n} \\
&= \frac{n^n-n-(n-1)^2}{(n-1)^2n^n} \\
&= \frac1{(n-1)^2}\left(1-\frac1{n^{n-1}}\right)-\frac1{n^n}.
\end{align*}
Therefore, for $n\geqslant3,$
\begin{align*}
\frac1{(n-1)^2} &=
\left(1-\frac1{n^{n-1}}\right)^{-1}\left(
\frac1{n^2}+\frac2{n^3}+\cdots+\frac{n-1}{n^n} +
\frac1{n^n}\right) \\ &=
\left(1-\frac1{n^{n-1}}\right)^{-1}\left(
\frac1{n^2}+\frac2{n^3}+\cdots+
\frac{n-2}{n^{n-1}}+\frac1{n^{n-1}}\right) \\ &=
\left(1+\frac1{n^{n-1}}+\frac1{n^{2n-2}}+\cdots\right)\left(
\frac1{n^2}+\frac2{n^3}+\cdots+\frac{n-3}{n^{n-2}}+\frac{n-1}{n^{n-1}}\right).
\end{align*}
This gives the expansion of $1/(n-1)^2$ in base $n.$
Taking $n=1000,$ we get:
$$
\frac1{999^2}=\left(1+\frac1{10^{2997}}+\frac1{10^{5994}}+\cdots\right)\left(
\frac1{10^6}+\frac2{10^9}+\cdots+
\frac{997}{10^{2994}}+\frac{999}{10^{2997}}\right),
$$
from which can be read off the decimal expansion of $1/998001.$ Similarly for $1/9801,$ $1/99980001,$ etc.
A bit shorter:
$$
1+2x+3x^2+\cdots+(n-2)x^{n-3} =
\frac{1-(n-1)x^{n-2}+(n-2)x^{n-1}}{(1-x)^2},
$$
$$
\therefore\ 1+\frac2n+\frac3{n^2}+\cdots+\frac{n-2}{n^{n-3}} =
\frac{1-n^{-(n-3)}+2n^{-(n-2)}-2n^{-(n-1)}}{(1-n^{-1})^2},
$$
\begin{multline*}
\therefore\
\frac1{n^2}+\frac2{n^3}+\frac3{n^4}+\cdots+\frac{n-2}{n^{n-1}}
= \frac{1-n^{-(n-3)}+2n^{-(n-2)}-2n^{-(n-1)}}{(n-1)^2} = \\
\frac{n^n-n^3+2n^2-2n}{(n-1)^2n^n}=\frac{n^n-n-n(n-1)^2}{(n-1)^2n^n}
= \frac1{(n-1)^2}\left(1-\frac1{n^{n-1}}\right)-\frac1{n^{n-1}},
\end{multline*}
\begin{gather*}
\therefore\ \frac1{(n-1)^2} =
\left(1-\frac1{n^{n-1}}\right)^{-1}\left(
\frac1{n^2}+\frac2{n^3}+\cdots+
\frac{n-2}{n^{n-1}}+\frac1{n^{n-1}}\right) \\
= \left(1+\frac1{n^{n-1}}+\frac1{n^{2n-2}}+\cdots\right)\left(
\frac1{n^2}+\frac2{n^3}+\cdots+\frac{n-3}{n^{n-2}}+
\frac{n-1}{n^{n-1}}\right).
\end{gather*}
