Third Squeeze Claim:If for some {$m$}, we have {$F_j=F_k$} for all {$j,k>m$}, then {$\mathcal S(F_0 F_1 F_2 F_3\ \ldots\ F_n)\longrightarrow 0$}. Proof: From the proof of the [First Squeeze Claim]], if {$C$} is represented infinitely often in the infinite sequence, then {$\mathcal S(F_0 F_1 F_2 F_3\ \ldots\ F_n)\longrightarrow 0$}, since {$\mathcal S(CG)\le \frac12\mathcal S(G)$}. The {$L,R,M$} are similar to each other. Here is {$R$} as an example. Suppose that for some {$n$}, we have {$F_j=R$} for all {$j>n$}. We have {$R^k(a,b,c)=(a,b,c+k(a+b)$}, or the point {$(\frac a{c+(k+1)(a+b)},\frac b{c+(k+1)(a+b)},\frac {c+k(a+b)}{c+(k+1)(a+b)})$}. Then either
In either case, {$\mathcal S(F_0 F_1 F_2 F_3\ \ldots\ F_n)\longrightarrow 0$} and the infinite traction converges to {$F_0 F_1 F_2 F_3\ \ldots\ F_m(0,0,1)$}. |