Conway’s Soldiers… part 2.

{\displaystyle \varphi ={\frac {{\sqrt {5}}-1}{2}}\approx 0.61803\,39887\ldots }
\sum _{{n=2}}^{{\infty }}\varphi ^{n}=1
The sigma sum

I will prove that the sigma sum is true given the value of phi, which is also the reciprocal of the golden ratio.

Remember how Φ^2 = 1 – Φ? The sigma sum states that 1 – Φ + Φ – Φ^2 + Φ^2 – Φ^3 + Φ^3 – Φ^4 + …, that means the powers of Φ keep cancelling themselves out, and if a number x is smaller than one, the limit as n goes to infinity of x^n approaches zero. And since Φ is less than one, Φ^∞ must converge to zero.

That means the final sum is 1 + Φ^∞ after all the cancellation, and since Φ^∞ converges to zero, the final sum is 1 + 0, or just one.


2 thoughts on “Conway’s Soldiers… part 2.

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