Why do we miss 8 in the decimal expansion of 1/81, and 98 in the decimal expansion of 1/9801? Why do we miss $8$ in the decimal expansion of $1/81$, and $98$ in the decimal expansion of $1/9801$? I've seen this happen that when you divide in a fraction using the square of any number with only nines in the denominator. Like in 
$$
  \frac{1}{9^2}=\frac{1}{81} = 
  0.01234567\!\underset{\uparrow}{}\!9
    01234567\!\underset{\uparrow}{}\!9
    01234567\dots\,,
$$
and in 
$$
  \frac{1}{99^2}=\frac{1}{9801} = 
  0.0001020304050607080910111213 \dots 9697\!\underset{\uparrow}{}\!99000102 \dots\;\,,
$$
the decimals go on predictably when suddenly in the first one you miss $8$, and in the second you miss $98$ and it keeps going on forever. How does this happen? Why do we miss numbers like $8$ in the decimal representation of $\frac{1}{9^2},$ or like $98$ in the decimal of $\frac{1}{99^2},$ or $998$ in $\frac{1}{999^2},$ or $9998$ in $\frac{1}{9999^2}\;$?

"Originally" there were 95, 96, 97, 98, 99, 100, 101, 102 ... but since there's only two digit positions for each of them, the one in front of 100, 101, and so forth carries over to the number in front of it and increases that preceding number by 1. This makes 99 into 100, and the one in front of that then carries over to 98 and makes it 99. And that's where the carries stop.
So the reason why exactly 98 is the one that is missing is that 98 is the largest two-digit number that can be increased by one without gaining another digit that would carry.

It may be instructive to see what happens if we look at some other "original" pattern than 0, 1, 2, 3, ..., 98, 99, 100, 101 ...
For example, $\frac{7}{9801}$ should produce

0, 7, 14, 28, ..., 84, 91, 98, 105, 112, 119 ...

but when carrying out the division the digits we get are:
7/9801 = 0.00071428...849199061320...

because of carries. Here one sees more clearly that 98 is not "missing", it is just increased by one, like the last two digits of the subsequent numbers in the sequence.
We can also look at
9/9801 = 0.00091827...819100091827...

Here, like in $\frac1{9801}$ there's a 99 in the original sequence, but it gets bumped up to 100 by the carry, and then the "original" 90 in turn gets bumped up to 91. Likewise
11/9801 = 0.0011223344556677890011223344...

There's two of every digit except 8 and 9! Again, what really happens is that the original 99 gets bumped to 100, which again bumps 88 to 89.
