Prove that $|z_{n+2}-z_{n+1}|\leq q\cdot|z_{n+1}-z_n| \implies (z_n)_{n\geq 1} \text{ converges}$ when $q\in (0,1)$

Question

Problem:

Let $q\in (0,1)$ and $(z_n)_{n\geq 1} \subset \mathbb{C}$ be a sequence. Prove:$$|z_{n+2}-z_{n+1}|\leq q\cdot|z_{n+1}-z_n| \implies (z_n)_{n\geq 1} \text{ converges}$$

My attempt:

$$|z_{n+2}-z_{n+1}|\leq q\cdot|z_{n+1}-z_n| \Longleftrightarrow \bigg |\frac{z_{n+2}-z_{n+1}}{z_{n+1}-z_n} \bigg | \leq q$$ If we want to remove the absolute value bars, we must note that $z_{n+2}\geq z_{n+1}$ and $z_{n+1}\geq z_n$ or $z_{n+2}\leq z_{n+1}$ and $z_{n+1}\leq z_n$, hence $(z_n)$ is either monotonically decreasing or monotonically increasing. Furthermore the quotient is either bounded above or bounded below, thus $(z_n)_n$ is convergent. $\Box$

Answer 1

Intuition/Hint:

Take $q=1/2$, $z_0=0$ and $z_1=1$. What can you say about $|z_n-z_{n-1}|$ for any given $n$? Further, what can you say about $|z_n-z_m|$ for any given $n$ and $m$?

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Here's a handy criterion (whose name I forgot) for convergence: $\{z_n\}$ converges if $$ \lim_{N \to \infty} \sup_{m,n>N} |z_n-z_m|=0. $$

Answer 2

You can’t say $z_{n+1} \ge z_n$; there is no “greater than” or “less than” in complex numbers.

Instead, note that by repeatedly applying the given formula, $|z_{n+1} - z_n| \le q |z_n - z_{n-1}| \le q^2 |z_{n-1}-z_{n-2} \ldots \le q^n |z_2 - z_1|$. Note that $|z_{n+1} - z_n|$ is the distance between the points $z_{n+1}$ and $z_n$ and is bounded by a geometric progression with ratio less than $1$. We now consider the two components separately, noting that difference in the $x$ coordinate is at most the distance between the two points. So we have $|x_{n+1} - x_n| \le q^n c$ for some constant $c$; this is bounded by a geometric series and so $x_n$ must converge (try proving this yourself)! The same holds for the $y$-coordinate, so $z_n$ converges.