MHB What is the pattern behind the repeating decimal in this interesting fraction?

[soroban]

\displaystyle\frac{1}{998,001} \;=\;0.\overline{000\,001\,002\,003\,004\,005\, \cdots\,996\,997\,999}\, \cdots

The decimal representation contains all the 3-digit numbers except 998
. . and the 2997-digit cycle repeats forever.

This is just one of a family of such fractions.
Can you determine the underlying characteristic?

 

[Opalg]

The pattern must be that this is the third member of a series starting

$\dfrac1{81} = \dfrac1{9^2} = 0.\overline{012345679}$,

$\dfrac1{9801} = \dfrac1{99^2} = 0.\overline{00\,01\,02\,03\,04\,05\cdots 96\,97\,99}$,

$\dfrac1{998\,001} = \dfrac1{999^2} = 0.\overline{000\,001\,002\,003\,004\,005\cdots 996\,997\,999}$.

You can sort of see why this works if you consider the binomial series

$(1-x)^{-2} = 1+2x+3x^2+4x^3+5x^4+\ldots$,

with $x = 1/10^k$. For example, if $k=1$ you get

$$\begin{aligned}\tfrac1{81} = 9^{-2} = (10-1)^{-2} &= \tfrac1{100}\bigl(1-\tfrac1{10}\bigr)^{-2} \\ &= \tfrac1{100}\bigl(1 + \tfrac2{10} + \tfrac3{100} +\tfrac4{1000} + \ldots\bigr).\end{aligned}$$

That gives you $\frac1{81} = 0.01234...$. The trouble starts when you have to try to justify why the 8 gets left out, and why the decimal starts to recur after the 9.

 

[mathworker]

The trouble starts when you have to try to justify why the 8 gets left out, and why the decimal starts to recur after the 9.

may be that's because after 9/10^8 we have 10/10^9=1/10^8...that implies
9/10^8+1/10^8=10/10^8=1/10^7...
which transforms or (eliminates to reproduce) 8/10^7 into 9/10^7 and recurring i am working on that...

 

[MarkFL]

There is additional discussion http://www.mathhelpboards.com/f9/1-998001-a-3403/.

 

[TheBigBadBen]

The pattern must be that this is the third member of a series starting

$\dfrac1{81} = \dfrac1{9^2} = 0.\overline{012345679}$,

$\dfrac1{9801} = \dfrac1{99^2} = 0.\overline{00\,01\,02\,03\,04\,05\cdots 96\,97\,99}$,

$\dfrac1{998\,001} = \dfrac1{999^2} = 0.\overline{000\,001\,002\,003\,004\,005\cdots 996\,997\,999}$.

You can sort of see why this works if you consider the binomial series

$(1-x)^{-2} = 1+2x+3x^2+4x^3+5x^4+\ldots$,

with $x = 1/10^k$. For example, if $k=1$ you get

$$\begin{aligned}\tfrac1{81} = 9^{-2} = (10-1)^{-2} &= \tfrac1{100}\bigl(1-\tfrac1{10}\bigr)^{-2} \\ &= \tfrac1{100}\bigl(1 + \tfrac2{10} + \tfrac3{100} +\tfrac4{1000} + \ldots\bigr).\end{aligned}$$

That gives you $\frac1{81} = 0.01234...$. The trouble starts when you have to try to justify why the 8 gets left out, and why the decimal starts to recur after the 9.

The answer, as it ends up, is simple. As you have hinted, we can write this as
$$\frac{1}{(10^n-1)^2}=\sum_{k=0}^{\infty}\frac{k}{10^{n(k+1)}}$$
To start, let's focus on the case of $\frac{1}{81}$, that is, the $n=1$ case. In this case, we have
1/81 =
.01 +
.002 +
.0003 + ...
which starts out fine. However, eventually we reach
...08 +
...009 +
...0010 +
...00011 + ...

now, the 1 from the 10 overlaps with the 9. As a result, there's a 1 that carries over to the eight, putting a 9 where there was an 8, and a 0 where there was a 9. In fact, because of the 1's in the 10's place of the numbers that follow, the sequence of numbers climbs onwards, until the next skip when that 1 changes to a 2. That is, the result of this sum after all the carries are accounted for is

...9012...

which gives us that characteristic "skip" two places back from $10^n$. Thus, the repeating decimal is $ 0.\overline{012345679}$, as we know. With a similar analysis, we find that there is always a skip of the $(10^n-1)^{th}$ $n$-digit number in the sequence. This skip is the inevitable result of the fact that our fraction hides the sum of an endlessly rising sequence of consecutive integers, not just the endless loop of $n$-digit integers that happen to come up in the digital representation.

For comparison, the number $ 0.\overline{0123456789}$ has the much less elegant (reduced) representation of $\frac{13717421}{1111111111}$.