CONTEXT
I need some help in solving an exercise from Conway's book "Functions of One Complex Variable I".
Let $D = \{z: |z| <1\}$ and show that $\mathcal F$ is normal iff there is a sequence $\{M_n\}$ of positive constants such that $\limsup \sqrt[n]{M_n} \leq 1$ and if $f(z) = \sum_{n=0}^\infty a_n z^n$, then $|a_n|\leq M_n$ for all $n$.
In this context normal means relatively compact.
WHAT I HAVE DONE
I managed to prove one of the two implications. Assume the existence of the sequence $\{M_n\}$. Then for each compact set $K\subset D$ there exists $r<1$ such that $K\subset \overline B(0,r) \subset D$, where $\overline B(0,r) = \{z:|z| \leq r\}$. Take $N$ large enough so that $M_n< \frac1{r}$. Then we have, for all $f\in \mathcal F$ and for all $z\in K$, \begin{eqnarray}|f(z)| &\leq& \sum_{n=0}^\infty |a_n| |z^n|=\sum_{n=0}^N M_n r^n + \sum_{n=N+1}^\infty \frac1{r}r^n\leq\\ &\leq&\max_{0\leq n\leq N} (M_n)\frac{1-r^{N+1}}{1-r}+\frac1{1-r}.\end{eqnarray} Then we can use Montel's Theorem to deduce relative compactness of $\mathcal F$.
As for the other implication, I guess one could use again Montel's Theorem to state the fact that, for any $r < 1$, for all $f\in F$ and for all $z\in \overline B(0,r)$ there is a constant $J_r$ such that $$|f(z)| \leq J_r.$$ Then $$|a_n| = \frac1{2\pi} \left|\int_{\gamma_r} \frac{f(z)}{z^{n+1}} dz\right|\leq \frac{J_r}{r^n},$$ but, since $J_r$ depends on $r$, I don't know how to use this bound to reach the required conclusion.
QUESTIONS
Is the first part of the exercise correct? Can you give me a hint on how to proceed to prove the inverse implication?
Thanks to Conrad's hint in comments, I should have been able to complete the exercise. Here is my attempt to the rest of the proof. Please have a look and feel free to correct/comment on it.
Suppose the assertion on the $a_n$'s is not true. Then for some $\delta>1$, there exist and sequence of positive integers $n_1,n_2,\dots$ and corresponding functions $f_{n_1},f_{n_2},\dots\in \mathcal F$, with $$f_k(z) = \sum_{h=1}^\infty a_{k,h} z^h,$$ such that $|a_{k,n_k}|>\delta^{n_k}$.
Normality implies that $\{f_k\}$ has a subsequence $\{f_{k_1},f_{k_2},\dots\}$ that converge to some $f \in H(D)$, with $$f(z) = \sum_{n=0}^\infty a_n z^n.$$
In particular we have, for each $m$, $$|a_{k_m,n_{k_m}}| \to |a_{n_{k_m}}|.$$
Since $|a_{k_m,n_{k_m}}|\geq \delta^{n_{k_m}}$, it follows that $$|a_{n_{k_m}}|\geq \delta^{n_{k_m}}.$$
Finally noting that $$\left|a_{n_{k_m}} \cdot \left(\frac{2}{1+\delta}\right)^{n_{k_m}}\right|\geq \left(\frac{2\delta}{1+\delta}\right)^{n_{k_m}}\not \to 0,$$ we conclude that the function $f(z)$ does not converge in $z= \frac{2}{1+\delta} <1$, a contradiction.