Let $(z_n)_{n \in \mathbb{N}} \subset \mathbb{C}$ and $q < 1$ so that
$$|z_{n+2} - z_{n+1}| \leq q|z_{n+1}-z_n|$$ for all $n \in \mathbb{N}$
How can one prove that
$$|z_{n+1} - z_n| \leq q^{n-1}|z_2-z_1|$$
for all $n \in \mathbb{N}$ and that the sequence converges?
This is what I have so far:
Let $(z_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{C}$ with the characteristics that $|z_{n+1} -z_n| \leq q^n$
My hypothesis is that $\forall z,l \in \mathbb{N}: |z_{n+l} - z_n| \leq q^n(\sum_{k=0}^{l-k}q^k)$
Let $z \in \mathbb{N}$. We prove the hypothesis by induction over $l \in \mathbb{N}$.
In case $l = 1$ we get the exact characteristics of the sequence $(z_n)_{n \in \mathbb{N}}:$ It is $|z_{n+1}-z_n| \leq q^n = q^n \big (\sum_{k=0}^0 q^k \big ).$
Let now $l$ be random/arbitrary and given the hypothesis for $l$ we now show it for $l+1$. Because of the triangle inequality it holds that
$$|z_{n+l+1}-z_n| \leq |z_{n+l+1}-z_{n+l}| + |z_{n+l}-z_n| $$
According to the hypothesis (which we use for the index $n+l$ instead of $n$) it follows that
$$|z_{n+l+1} - z_{n+l}| \leq q^{n+l} = q^nq^l$$
With that the first addend has been estimated. For the second one we use the induction hypothesis and get
$$|z_{n+l+1}-z_{n+l}| + |z_{n+l}-z_n| \leq q^n q^l+q^n \big (\sum_{k=0}^{l-1}q^k \big ) = q^n \big (q^l + \sum_{k=0}^{l-1}q^k \big ) = q^n \big (\sum_{k=0}^{(l+1)-1} q^k \big ) $$
Hence it is $|z_{n+l+1} - z_n| \leq q^n \big (\sum_{k=0}^{(l+1)-1} q^k\big)$ which we wanted to prove.
I feel like this proof is incomplete or wrong (convergence missing too). Is there another way to do it?
You are very nearly there. What you want to show is that the sequence is Cauchy, so write out that definition and verify.
Hint: For $m,l > n$, show that