= Negative-parameter square-root iteration changes sign infinitely often
Let $c=-d<0$ and suppose the <recurrence relation> $a_{n+1}=(a_n-d/a_n)/2$ is defined with no zero iterates. An eventually constant sign would make $b_n=|a_n|$ satisfy $0<b_{n+1}=(b_n-d/b_n)/2<b_n/2$ on a tail. Thus $b_n\to0$, although positivity of $b_{n+1}$ requires $b_n^2>d$ throughout that tail. This contradiction proves infinitely many sign changes. The nonzero assumption matters because a zero iterate makes the next step undefined.
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