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

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.