prove that $\lim\sup_{r \to \infty} \frac{M_{f}(r)}{r^n} = \infty$ I need some help with this question.
Let $f(z)$ an entire function, with infinite number of zeros.
I want to prove that $\lim\sup_{r \to \infty} \frac{M_{f}(r)}{r^n} = \infty$
The definition of $M_{f}(r)$ is $\sup_{|z|=r}|f(z)|$, $r\in(0,R)$ when $f$ analytic in ${|z|<R}$
Assume that $f(z) \not\equiv 0$.
Thanks.
 A: Suppose that $\limsup_{r \rightarrow \infty}\frac{M_f(r)}{r^n}<\infty$.
This implies that there exists a constant $M$ such that for all $r$ sufficiently large, 
$$|f(re^{i\theta})| \leq M r^n$$ for every $\theta$.
Write $f(z)= \sum_{k=0}^{\infty}a_kz^k$. Fix $r$ large. By Taylor's theorem,
$$a_k=\frac{1}{2\pi i}\int_{\{|z|=r\}} \frac{f(z)}{z^{k+1}}dz$$
and thus
$$|a_k| \leq \frac{2\pi r Mr^n}{2\pi r^{k+1}} = Mr^{n-k}.$$
This holds for all $r$ sufficiently large, so if $k>n$, we find $a_k=0$, by letting $r \rightarrow \infty$. Thus, $f$ is a polynomial of degree at most $n$. Since $f$ is not identically zero, then $f$ has a finite number of zeros.
A: The Riemann sphere is compact, so the zeroes have to accumulate somewhere.  Since the zeros in the plane are a discrete set (otherwise $f(z) = 0$ everywhere), they must accumulate at $\infty$.  Then $f$ has an essential singularity at $\infty$ and the desired result follows.
(More generally, any non-polynomial entire function has an essential singularity at $\infty$.)
