# Show the complex sequence converges to $\sqrt{2}$

Let $f(z) = \frac{z+2}{z+1}$ for $z\neq-1$. If $z_1=i$ and $z_{n+1}=f(z_n)$ for every $n\in\mathbb{N}$, then show that the sequence $(z_n)^\infty_{n=1}$ converges to $\sqrt{2}$.

I was given the hint to show first that the sequence is Cauchy, and from there to show that it converges to $\sqrt{2}$, but I do not know how to show a recursively defined sequence is Cauchy.

I tried showing that $|z_n-z_{n+1}|<\epsilon$ but that doesn't seem to prove the requirements for a Cauchy sequence. I also tried showing $|z_n-z_{n+k}|<\epsilon$ but that became horribly messy.

The map $T(z) = \frac{z+2}{z+1}$ is a contraction on the set $\{z \in \mathbb{C}: \, |z+1| \ge \sqrt{2}, \, \mathrm{Re}[z] \ge 0\}$ because it maps this set to itself and because
$$|T(z) - T(w)| = \Big| \frac{w-z}{(w+1)(z+1)} \Big| \le \frac{1}{2}|w-z|.$$
The Banach fixed point theorem implies that the sequence $z, T(z),T(T(z)),...$ converges to the unique fixed point $T(z) = z$ on that set, which you can solve to be $\sqrt{2}.$
• @krakenwagon Using the inequality $|T(z) - T(w)| \le \frac{1}{2}|w-z|$ you can show inductively that $|T^{(m+1)}(z) - T^{(m)}(z)| \le \frac{1}{2^m}|w-z|$ for all $m$, and the triangle inequality implies that it's Cauchy Jan 29, 2017 at 3:11