[madcaptenor]

.123456... = x + 2 x^2 + 3 x^3 + ... with x = 1/10.

Then you have (x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...) + (x^3 + x^4 + x^5 + x^6 + ...) (count the number of occurrences of each power of x^n on the right-hand side)

and from the sum of a geometric series the RHS is x/(1-x) + x^2/(1-x) + x^3/(1-x) + ..., which itself is a geometric series and works out to x/(1-x)^2. Then put in x = 1/10 to get 10/81.

Now 0.987654... = 1 - 0.012345... = 1 - (1/10) (10/81) = 1 - 1/81 = 80/81.

[gowld]

Don't need the clutter of infinite series and polynomials:

    1/9 = 0.1111...

    1/81 = 1/9 * 1/9 = 0.111... * 0.111... =

    Sum of:
       0.0111...
       0.00111...
       0.000111...
       ...
    
    =  0.012345...

[GuB-42]

Isn't it essentially the same thing, but less formal

0.1111... is just a notation for (x + x^2 + x^3 + x^4 + ...) with x = 1/10

1/9 = 0.1111... is a direct application of the x/(1-x) formula

The sum of 0.0111... + 0.00111... ... = 0.012345... part is the same as the "(x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...)" part (but divided by 10)

And 1/81 = 1/9 * 1/9 ... part is the x/(1-x)^2 result

[madcaptenor]

This is better than my answer, at least if you can get your brain to interpret it in base b. In that case the first two lines would become

  1/(b-1) = 0.1111...
  1/((b-1)^2) = 1/b * 1/b = 0.111... * 0.111... =

[gus_massa]

I don't know who downvoted this, but it's correct.

The use of series is a little "sloppy", but x + 2 x^2 + 3 x^3 + ... has absolute uniform convergence when |x|<r<1, even more importantly that it's true even for complex numbers |z|<r<1.

The super nice property of complex analysis is that you can be almost ridiculously "sloppy" inside that open circle and the Conway book will tell you everything is ok.

[I'll post a similar proof, but mine use -1/10 and rounding, so mine is probably worse.]