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.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 + ...) +

.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.

gus_massa 1y ago
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.]
gowld 1y ago
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...
- madcaptenor 1y ago
  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... =
- GuB-42 1y ago
  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

gus_massa 1y ago

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.]

gowld 1y ago

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...

madcaptenor 1y ago
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... =
GuB-42 1y ago
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