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 ?
 A: 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 $1$.
A: take $z_n = c$ a constant. Surely the inequality is satisfied and the limit is $C$. So the limit could be any complex number. Maybe you are supposed to show the sequence converges rather than compute its limit.
A: Let $n\in \Bbb N$ and $L =z_n$,  then we have
$$ \lim_{m\to \infty}|L-z_m|=\lim_{m\to \infty}|z_n-z_m| < \lim_{m\to\infty} \frac{1}{1+|n-m|}=0.$$
It follows that the sequence $(z_m)_{m\in \Bbb N}$ converges to $z_n$. Since $(z_m)_{m\in \Bbb N}$ is a convergent sequence its limit is unique. However, we know that $(z_m)_{m\in \Bbb N}$ converges to $L=z_n$. It follows that $L=z_0=z_1=\ldots$, i.e. $(z_m)_{m\in \Bbb N}$ is a constant sequence.
