2 ms·

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...

• 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