TITLE: Prove that $z_{n+1} = \frac12 \left(z_n +\frac{1}{z_n} \right) $ converges to $1$
QUESTION [6 upvotes]: For a complex sequence $\{z_n\} $ defined by the recurrence relation $$z_{n+1}=\frac12  \Bigg(z_n +\frac{1}{z_n} \Bigg) $$ where $\arg z_n \in (-\frac{\pi}{2},\frac{\pi}{2})$, prove that it converges to $1$.

REPLY [14 votes]: From the recursion relation we get, $$\left(\frac{z_{n+1}+1}{z_{n+1}-1}\right)=\left(\frac{z_{n}+1}{z_{n}-1}\right)^2, \forall n\geq1 $$ So iteratively, we get, $$\left(\frac{z_{n+1}+1}{z_{n+1}-1}\right)=\left(\frac{z_{0}+1}{z_{0}-1}\right)^{2^{n+1}}$$ Rearranging, we get, $$z_{n+1}=\frac{1+w^{2^{n+1}}}{1-w^{2^{n+1}}}$$ where $$w=\frac{z_0-1}{z_0+1}$$ Clearly, $|w|<1$ and $\arg{w}\in[-\pi/2, \pi/2]$ (I'm omitting the proofs but they are not difficult). Hence $$\lim_{n\rightarrow \infty}w^{2^{n+1}}=0 e^{j0}=0
$$ Thus $$\lim_{n\rightarrow \infty}z_n=1 \hspace{0.6cm}\Box$$