Newton iteration for a square root ($a_{n+1}=\frac{1}{2}(a_n+c/a_n)$)

For $c>0$ and $a_0>0$, the displayed recurrence relation stays positive and satisfiesAfter the first step, $a_n^2\ge c$, so $a_{n+1}-a_n=(c-a_n^2)/(2a_n)\le 0$. The bounded monotone sequence theorem gives a limit of a sequence $L\ge 0$; the inequality $L^2\ge c>0$ makes $L$ positive, and passage to the limit of a sequence in the recurrence relation gives $L^2=c$. This proves existence of the positive square root without assuming it in the construction. Once $s=\sqrt{c}$ is known, the exact error identity $a_{n+1}-s=(a_n-s{)}^2/(2a_n)$ explains the eventual quadratic convergence.

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 $b_n\to 0$, 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.