Yesterday, I posted a proof on why the sigma sum for phi was true, but I kind of rushed it, so I will be going into details.

If you can do a little bit of “hard maths”, then it is easy to prove that:

is true.

That means that the sum is Φ^2 + Φ^3 + Φ^4 + Φ^5 + Φ^6 + …, but since Φ^2 = 1 – Φ, that means that Φ^3 = Φ^2 * Φ, or Φ-Φ^2. If we keep multiplying, we begin to see a pattern here, where the exponent keeps going up by one, and in the full sum, there is, for any positive integer k, a Φ^k and a -Φ^k, which implies that they can be cancelled out. But there is no term to cancel the one in Φ^2, so one is left, and Φ^∞, since Φ is smaller than one, the limit as x approaches infinity of Φ^x converges to 0. That means the final sum is 1-0, or just 1.