3 ms·

Very nice indeed, but you can do it in binary, where the series is 0.01 + 0.0001 + 0.000001 + ... = 0.010101010101... Multiplying the right hand side by 2 g

• michael_nielsen 17y ago
  The advantage of my proof is that it generalizes to give the sum of 1/n + 1/n.n + ... Just replace 4 everywhere by n, and 3 by n-1: 1/n + 1/n.n + ... = 0.1111... (in base n) Multiply by (n-1) to get: 0.(n-1)(n-1)(n-1)... = 1 So the original sum must have been 1/(n-1). This is not so easy in binary.