Meromorphic function with infinitely many poles must have essential singularity at point of infinity?

Question

Question: Suppose $f: \mathbb{C}\setminus P \to \mathbb{C}$ is meromorphic on the complex plane, where $P:=\{z_n: n\in\mathbb{N}\}$ is a set of infinitely many isolated poles with $\infty$ as their only limit point on the extended complex plane. With this condition, can we conclude $f$ has essential sigularity at $\infty$?

I'm know if we require $f$ to be meromorphic on the extended complex plane, then there are at most finitely many poles, where $\infty$ can be either a removeable singularity or a pole. Since then $f$ must be holomorphic on $|z|>R$ for some $R>0$ and its complement is a compact set. But for a function to be meromorphic on $\mathbb{C}$, it doesn't matter how it behaves at $\infty$, and nothing prevents it from having a sequence of poles diverging to $\infty$. Hence the question. Thanks.

Answer 1

If a meromorphic function $f(z)$ does not have essential singularity at $z=\infty$, then the principal part of the Laurent expansion of $f(w^{-1})$ at $w=0$ only contains finitely many terms. That is, there is some $R>0$ such that when $|w|<R^{-1}$, there is some $N\in\mathbb N$ such that

$$ f(w^{-1})=\sum_{n\ge1}b_nw^n+a_0+a_1w^{-1}+\dots+a_Nw^{-N}. $$

In other words, for $|z|>R$, there is

$$ f(z)=\sum_{n\ge1}b_nz^{-n}+a_0+a_1z+\dots+a_Nz^N. $$

This indicates that $f$ has no poles other than $z=\infty$ for $|z|>R$. Furthermore, because $f$ is meromorphic, it can only possess finitely many poles in $|z|<R$, so we conclude that

Proposition 1: If $f$ is meromorphic in $\mathbb C$ and at $\infty$, then it can only possess finitely many poles.

Because essential singularity needs to be isolated, the contrapositive of this proposition is

Proposition 2: If $f$ is meromorphic in $\mathbb C$ and has infinitely many poles, then $z=\infty$ is a singularity of $f$ that is not isolated.