Spacy Colors Lemma (Posted on 2014-12-06)

   ak ≥ ak+1 > 0 for all k≥1.

   rk = ak                        for k = 1,

      = ak*√[ 1 - tk*tk ]          for k > 1,

   where tk = ak/(2*rk-1).

	Submitted by Bractals
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Proof that tk∈(0,1) for all k>1: t2 = a2/(2*r1) = a2/(2*a1) ≤ a1/(2*a1) = 1/2 < sqrt[ 1/2 ] tk < sqrt[ 1/2 ] ⇒ tk*tk < 1/2 ⇒ 1/2 < 1 - tk*tk ⇒ sqrt[ 1/2 ] < sqrt[ 1 - tk*tk ] ⇒ ak*sqrt[ 1/2 ] < ak*sqrt[ 1 - tk*tk ] ⇒ ak*sqrt[ 1/2 ] < rk ⇒ ak+1*sqrt[ 1/2 ] < rk ⇒ ak+1/(2*rk) < 1/(2*sqrt[ 1/2 ]) ⇒ tk+1 < sqrt[ 1/2 ] Therefore, tk < sqrt[ 1/2 ] < 1 for all k>1. Clearly, tk > 0 for all k>1 unless a tk-1 = 1. Therefore, tk∈(0,1) for all k>1. QED