# 1/9801

A couple of days ago, I remembered the following fun fact from my high-school days in Romania:
$! \frac{1}{9801}=0,(000102030405\cdots939495969799).$

First, note that I am using the European notation for periodic decimal expansions (or repeating decimals). What you get is a periodic expansion, where all the two-digit numbers from 00 to 99, except 98, appear in the periodic part. Ok, there seems to be some magic going on here, let us try to understand what is happening.
First, note that $9801 = 99^2$, so probably there’s nothing special about the 2 in there. Indeed, try
$! \frac{1}{9^2} = \frac{1}{81} = 0,(012345679).$

Good, let us go ahead and generalize this a little more. The 9 in there is the number of fingers you are left with when you lose one, so the same should be true for creatures anatomically different (by the way, there is ice on Mercury, and maybe more !):
$! \frac{1}{\text{E1}_{16}} = \frac{1}{\text{15}_{16}^2} = 0,(012345679\text{ABCDF})_{16}.$

Before stating and proving a general result, let us discuss the easiest case (for us humans), 1/81. The idea comes from [cgo] and the starting point is the formula
$! \sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2},$

which is the power series expansion of $(1-x)^{-2}$ around $0$. For $x=1/10$, we get
$! \begin{array}{rcl} \frac{1}{9^2} &=& \frac{1}{10}^2 \left[ 1 + \frac{2}{10} + \frac{3}{10^2} + \cdots \right] \\&=& \left[ \frac{0}{10} + \frac{1}{10^2} + \frac{2}{10^3} + \cdots + \frac{7}{10^8} + \frac{8}{10^9} + \right. \\&& \qquad \left. + \frac{9}{10^{10}} + \frac{10}{10^{11}} + \frac{11}{10^{12}} + \frac{12}{10^{13}} + \cdots \right]. \end{array}$

At this point, the expression above starts to look like a decimal expansion – the problem is that some of the numerators in the brackets are not digits! To get around this, write
$! \begin{array}{rcl}\frac{10}{10^{11}} &=& \frac{1}{10^{10}} + \frac{0}{10^{11}} \\ \frac{11}{10^{12}} &=&\frac{1}{10^{11}} + \frac{1}{10^{12}} \\ \frac{12}{10^{13}} &=& \frac{1}{10^{12}} + \frac{2}{10^{13}} \\ &\cdots \end{array}$
to get
$!\begin{array}{rcl} \frac{1}{9^2} &=& \left[ \frac{0}{10} + \cdots + \frac{7}{10^8} + \frac{8}{10^9} + \frac{9}{10^{10}} + \left( \frac{1}{10^{10}} + \frac{0}{10^{11}} \right ) + \right. \\ && \qquad \left. + \left( \frac{1}{10^{11}} + \frac{1}{10^{12}}\right ) + \left( \frac{1}{10^{12}} + \frac{2}{10^{13}} \right ) + \cdots \right] \\ &=& \left[ \frac{0}{10} + \cdots + \frac{7}{10^8} + \frac{8}{10^9} + \left(\frac{9}{10^{10}} + \frac{1}{10^{10}}\right ) + \left( \frac{0}{10^{11}} + \frac{1}{10^{11}} \right ) + \right. \\ && \qquad \left. + \left( \frac{1}{10^{12}}+\frac{1}{10^{12}} \right ) + \frac{2}{10^{13}} + \cdots \right] \\ &=& \left[ \frac{0}{10} + \cdots + \frac{7}{10^8} + \frac{8}{10^9} + \frac{10}{10^{10}} + \frac{1}{10^{11}}+\frac{2}{10^{12}}+\cdots \right] \\ &=& \left[ \frac{0}{10} + \cdots + \frac{7}{10^8} + \frac{8}{10^9} + \left( \frac{1}{10^{9}} +\frac{0}{10^{10}} \right)+ \frac{1}{10^{11}}+\frac{2}{10^{12}}+\cdots \right] \\ &=& \left[ \frac{0}{10} + \cdots + \frac{7}{10^8} + \frac{9}{10^9} + \frac{0}{10^{10}} + \frac{1}{10^{11}}+\frac{2}{10^{12}}+\cdots \right],\end{array}$

which is the announced formula. The general result can be stated as follows.

Proposition 1. Let $b \geq 3$ be a fixed basis. Then, when everything is written in basis $b$, for any $n \geq 1$, one has
$! \begin{array}{rc} \frac{1}{\overline{b-1} \, \overline{b-1}\cdots\overline{b-1} \, \overline{b-1}^{2}} = 0,(&00\cdots00 \\&00\cdots01 \\&00\cdots02 \\&\qquad \vdots\\ &\overline{b-1} \, \overline{b-1}\cdots\overline{b-1} \, \overline{b-4} \\ &\overline{b-1} \, \overline{b-1}\cdots\overline{b-1} \, \overline{b-3} \\ &\overline{b-1} \, \overline{b-1}\cdots\overline{b-1} \, \overline{b-1}), \end{array}$

where each group containing $\cdots$ in the expression above has $n$ digits and $\overline{x}$ denotes the digit $0 \leq x < b$ in basis $b$. Proof: We shall start from the RHS of the equation and work our way towards the LHS. Recall that a periodic decimal expansion in base $b$ can be written as a fraction as follows:
$! 0,(\overline{x_{k-1}} , \overline{x_{k-2}} , \overline{x_{k-3}} \cdots \overline{x_{1}} , \overline{x_0} ) = \frac{\sum_{i=0}^{k-1} \overline{x_i} b^i}{ (b-1) \sum_{i=0}^{k-1} b^i}.$

First, let us focus on the case $n=1$ and treat the general case at the end of the proof. In this particular case, the RHS of the equation in the statement is equal to
$! 0,(012\cdots \overline{b-3} , \overline{b-1}) = \frac{1+\sum_{i=0}^{b-2} (b-2-i)b^i}{(b-1)\sum_{i=0}^{b-2}b^i},$

where the 1 in front of the sum stands for the fact that $\overline{b-2}$ is replaced by $\overline{b-1}$ as the last digit of the period. Using the formulas
$! \begin{array}{rcl} \sum_{i=0}^k b^i &=& \frac{b^{k+1}-1}{b-1}\\ \sum_{i=0}^k i b^i &=& \frac{b+b^{k+1}(kb-k-1)}{(b-1)^2}, \end{array}$

we can write
$! \begin{array}{rcl} 1+\sum_{i=0}^{b-2} (b-2-i)b^i &=& 1+(b-2)\frac{b^{b-1}-1}{b-1} – \\ && \quad – \frac{b+b^{b-1}[(b-2)b-(b-2)-1]}{(b-1)^2} \\ &=& \frac{b^{b-1}-1}{(b-1)^2}, \end{array}$

from which the conclusion easily follows. For the general case, note that the following decimal expansion holds for groups $\widetilde y_i$ of $n$ digits:
$! 0,(\widetilde{y_{k-1}} , \widetilde{y_{k-2}} , \widetilde{y_{k-3}} \cdots \widetilde{y_{1}} , \widetilde{y_0} ) = \frac{\sum_{i=0}^{k-1} \widetilde{y_i} b^{ni}}{(b-1)\sum_{i=0}^{nk-1} b^i}.$

The proof follows then the same steps as above.

References

[cgo] A. Cheer and D. Goldston, Explaining the decimal expansion of 1/81 using calculus. Mathematics and Computer Education, 25-3 (1991).