# Normal Family of Entire Functions, Convergence

I am looking for hints for this old prelim exam question.

Let $$\mathcal{F}$$ be a normal family of entire functions, and let $$A_n=\sup\left\{|a_n|:f(z)=\sum_{j=0}^{\infty}a_jz^j,f\in\mathcal{F}\right\}$$ for $$n=0,1,2,\ldots.$$ Show that $$\sum_{n=0}^{\infty}A_nr^n<\infty$$ for all $$r>0.$$

My first step is to recognize that $$A_n=\frac{1}{n!}\sup_{f\in\mathcal{F}}|f^{(n)}(0)|.$$ For each $$n,$$ there exists a sequence of functions $$(f_{n,k})_{k=1}^{\infty}$$ in $$\mathcal{F}$$ such that $$\lim_{k\to\infty}\frac{\left|f_{n,k}^{(n)}(0)\right|}{n!}=A_n.$$ Now since $$\mathcal{F}$$ is a normal family, the sequence $$(f_{n,k})_{k=1}^{\infty}$$ admits a subsequence $$\left(f_{n,k_j}\right)_{j=1}^{\infty}$$ such that there exists a function $$g_n\colon \mathbb{C}\to\mathbb{C}$$ such that $$f_{n,k_j}\xrightarrow[j\to\infty]{}g_n$$ uniformly on compact sets. Then it can be shown that $$g_n$$ is entire for each $$n,$$ and $$f_{n,k_j}^{(m)}\xrightarrow[j\to\infty]{}g_n^{(m)}$$ uniformly on compact sets for all orders $$m\in \mathbb{N},$$ in particular $$f_{n,k_j}^{(n)}\xrightarrow[j\to\infty]{}g_n^{(n)}$$ uniformly on compact sets. It follows that $$\frac{\left|g_n^{(n)}(0)\right|}{n!}=A_n.$$ My hope is to use the functions $$g_n$$ find an entire function $$h$$ with the property that $$h^{(n)}=g_n^{(n)}$$ for all orders $$n,$$ but don't see how to do this. Does this seem to be a good strategy, or is there a better approach I am missing? I feel that I am not making enough use of the fact that $$\mathcal{F}$$ consists of entire functions - is there a way to use this fact here?

$$\cal F$$ is uniformly bounded on compact sets, so that $$M_R = \sup \{ |f(z)| : |z| \le R, f \in \cal F \}$$ is finite for every $$R> 0$$. Using Cauchy's integral formula, this gives a uniform upper bound for $$|a_n|$$, and therefore an upper bound for $$A_n$$.
$$A_n \le \frac{M_R}{R^n}$$.
Using this bound, you can prove the convergence of $$\sum_{n=0}^{\infty}A_nr^n$$ for all $$r > 0$$, for example with the root test, or by comparison with the geometric series.
Choose $$R= 2r$$. Then $$A_n r^n \le \frac{M_R}{2^n}$$.