Also, adding 123456789 to itself eight times on an abacus is a nice exercise, and it's easy to visually control the end result.

"Why include a script rather than a proof? One reason is that the proof is straight-forward but tedious and the script is compact.

A more general reason that I give computational demonstrations of theorems is that programs are complementary to proofs. Programs and proofs are both subject to bugs, but they’re not likely to have the same bugs. And because programs made details explicit by necessity, a program might fill in gaps that aren’t sufficiently spelled out in a proof."

0.987654321/0.123456789 = (1.11111111-x)/x = 1.11111111/x - 1 where x = 0.123456789

You can aproxímate 1.11111111 by 10/9 and aproxímate x = 0.123456789 using y = 0.123456789ABCD... = 0.123456789(10)(11)(12)(13)... that is a number in base 10 that is not written correctly and has digits that are greater than 9. I.E. y = sum_i>0 i/10^i

Now you can consider the function f(t) = t + 2 t^2 + 3 t^3 + 4 t^4 + ... = sum_i>0 i*t^i and y is just y=f(0.1).

And also consider an auxiliary function g(t) = t + t^2 + t^3 + t^4 + ... = sum_i>0 1*t^i . A nice property is that g(t)= 1/(1-t) when -1<t<1.

The problem with g is that it lacks the coefficients, but that can be solved taking the derivative. g'(t) = 1 + 2 t + 3 t^2 + 4 t^3 + ... Now the coefficients are shifted but it can be solved multiplying by t. So f(t)=t*g'(t).

So f(t) = t * (1/(1-t))' = t * (1/(1-t)^2) = t/(1-t)^2

and y = f(0.1) = .1/.9^2 = 10/81

then 0.987654321/0.123456789 ~= (10/9-y)/y = 10/(9y)-1 = 9 - 1 = 8

Now add some error bounds using the Taylor method to get the difference between x and y, and also a bound for the difference between 1.11111111 an 10/9. It shoud take like 15 minutes to get all the details right, but I'm too lazy.

(As I said in another comment, all these series have a good convergence for |z|<1, so by standards methods of complex analysis all the series tricks are correct.)

In general, sum(x^k, k=1…n) = x(1-x^n)/(1-x).

Then sum(kx^(k-1), k=1…n) = d/dx sum(x^k, k=1…n) = d/dx (x(1-x^n))/(1-x) = (nx^(n+1) - (n+1)x^n + 1)/(1-x)^2

With x=b, n=b-1, the numerator as defined in TFA is n = sum(kb^(k-1), k=1…b-1) = ((b-2)b^b + 1)/(1-b)^2 = ((b-2)b^b + 1)/(1-b)^2.

And the denominator is:

d = sum((b-k)b^(k-1), k=1..b-1) = sum(b^k, k=1..b-1) - sum(kb^(k-1), k=1..b-1) = (b-b^b)/(1-b) - n = (b^b - b^2 + b - 1)/(1-b)^2.

Then, n-(b-1) = (b^(b+1) - 2b^b - b^3 + 3b^2 - 3b +2)/(1-b)^2.

And d(b-2) = the same thing.

So n = d(b-2) + b - 1, whence n/d = b-2 + (b-1)/d.

We also see that the dominant term in d will be b^b/(1-b)^2 which grows like b^(b-2), which is why the fractional part of n/d is 1 over that.

I disagree with the author that a script works as well as a proof. Scripts are neither constructive nor exhaustive.

    ┌───┬───┬───┐
    │ 7 │ 8 │ 9 │
    ├───┼───┼───┤
    │ 4 │ 5 │ 6 │
    ├───┼───┼───┤
    │ 1 │ 2 │ 3 │
    ├───┼───┼───┤
    │ 0 │ . │   │
    └───┴───┴───┘

987,654,321 + 123,456,789 = 1,111,111,110

1,111,111,110 + 123,456,789 = 1,234,567,899 \approx 1,234,567,890

So 987,654,321 + 2 x 123,456,789 \approx 10 x 123,456,789

Thus 987,654,321 / 123,456,789 \approx 8.

If you squint you can see how it would work similarly in other bases. Add the 123... equivalent once to get the base-independent series of 1's, add a second time to get the base-independent 123...0.

  > 987654320 / 123456790
  8.0

   ( 987654321 - 1 )
   -----------------  = 8
   ( 123456789 + 1 )

  1> (/ (poly 4 '(3 2 1)) (poly 4 '(1 2 3)))
  2.11111111111111
  2> (/ (poly 4 '(3 2 0)) (poly 4 '(1 2 4)))
  2.0

For base 2, the ratio is 1/1. When we apply the correction, we get (1 - 1) / (1 + 1) = 0, which is 2 - 2.