Evaluating a sequence of complex numbers: $|z_n - z_m| < \frac1{1+ |n-m|}$

Let $\{z_n\}$ be sequence of complex numbers such that

$$ |z_n - z_m| < \frac{1}{1+ |n-m|} $$ for all $n,m$

Given this information, can we compute $\lim_{n \to \infty} z_n $?

Attempt:

For sure the sequence is bounded since the RHS is always $\leq 1 $. IF $z_m = 0$, then we have

$$ |z_n | < 1 $$ for all $n$. I claim that the limit is $1$. How can I prove this ?

Consider $$|z_n - z_m| < \frac{1}{1+|n-m|}$$ for some fixed $m$ and arbitrarily large $n$. This immediately leads to the conclusion that $$\lim_{n \to \infty} z_n = z_m.$$ But this is also true for $m' \ne m$, so we are forced to conclude that $z_m = z_{m'}$ for all $m' \ne m$ if such a sequence exists. Therefore, the only sequence that satisfies the given conditions is a constant sequence. Since the choice $z_n = c$ for all $n$ satisfies the inequality, we conclude that it is not possible to show that the limit of such a sequence is necessarily $1$.