Why is $\frac{987654321}{123456789} = 8.0000000729?!$ Many years ago, 
I noticed that $987654321/123456789 = 8.0000000729\ldots$.
I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!
My life has gone downhill since then:)
My questions are:


*

*Why is this so?

*What happens beyond the "$729$"?

*What happens in bases other than $10$?
 A: $$729=9^3$$
$$66339=9^3\cdot 91$$
$$6036849=9^3\cdot 91^2$$
$$...$$
$$987654321/123456789=8+9^3\cdot 10^{-10}\cdot\displaystyle\sum_{n=0}^{\infty}(91\cdot 10^{-10})^n$$
A: Let 
$$S_n(a)=1 +2a+\ldots +na^{n-1}=\frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$
$$T_n(a)=a^{n-1}+2a^{n-2}+\ldots +n=a^{n-1}S_n(a^{-1}).$$
Then
$$
\frac{S_n(a)}{T_n(a)}=\frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$
For $a=10,n=9$ we have
$$
\frac{S_n(a)}{T_n(a)}\approx\frac{8\cdot 10^{10}+1}{10^{10}}.
$$
A: Just to add to the excellent answers above, some examples:
${987654321\,/\,123456789}\approx 8.00000007290000066339$
${{87654321}_9\,/\,{12345678}_9}\approx {7.000000628000056238}_9$
${{7654321}_8\,/\,{1234567}_8}\approx {6.0000052700046137}_8$
${{654321}_7\,/\,{123456}_7}\approx {5.00004260036036}_7$
${{\mathrm{fedcba987654321}}_{16}\,/\,{\mathrm{123456789abcdef}}_{16}}\approx {\mathrm{e.0000000000000d2f00000000000c693f}}_{16}$
A: I think @BorisNovikov's answer is the best here, but I had to do some work to understand it; this is my attempt at clarifying his answer, hopefully this can help others too: 
In the following, we adopt the notation $123456789 = [ 1, 2, ..., B-1 ]_B$ in base $B=10$. 
Boris introduced the following sum:
$$
  S_d(B) = [d, d-1, ..., 1]_B = 1 + 2B + ... + dB^{d-1} = \sum_{k=1}^d kB^{k-1} = \frac{dB^{d+1} - (d+1)B^d + 1}{(B-1)^2}
$$
where $d$ is the number of digits, $B$ the base, and the final formula is obtained by derivation of the usual geometric sum. 
From there, it is easy to see that:
$$
  [1, 2, ..., d]_B = d + (d-1)B + ... + 1B^{d-1} = B^{d-1} S_d(1/B)
$$
and putting both results together, we have:
$$
  \frac{ [B-1, B-2, ..., 1]_B }{ [1, 2, ..., B-1]_B } = \frac{ S_{B-1}(B) }{ B^{B-2} S_{B-1}(1/B) } = \frac{1 + (B-2)B^B}{B^B - B(B-1) - 1}
$$
Clearly both the numerator and denominator are dominated by the terms in $B^B$, so this fraction is equivalent to $B-2$ as $B$ tends to infinity, and we can see that the exact value is very close already for $B=10$ (see also @Mark Adler's post for other examples). Beautiful, isn't it? :) 
A: I think one should also think about how this is related to 9 being before 10. Notice the two following striking facts:
$12345679*8=98765432$ and $12345679*9=111111111$
A: In base $n$ the numerator is $$p = n^{n-1} - \frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = \frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$
Note that $p = (n-2)q + n-1$ and for the quotient we get
\begin{align}
\frac{p}{q} &= n-2 + \frac{(n-1)^3}{n^n} \frac{1}{1 - \frac{n^2-n+1}{n^n}} \\
&= n-2 + \frac{(n-1)^3}{n^n} \sum_{k=0}^{\infty} \left(\frac{n^2-n+1}{n^n}\right)^k.
\end{align}
Indeed for $n=10$ this is
$$\frac{987654321}{123456789} = 8 + \frac{729}{10^{10}}\sum_{k=0}^{\infty}\left(\frac{91}{10^{10}}\right)^k
$$
