Question: Is it true that if $\{f_n\}\subset A(\mathbb{\Omega})$ is normal then for some sequence $\{m_n\}$, $m_n\rightarrow\infty$, a subsequence $\{f_n^{m_n}\}$ is normal.
My Thoughts: Since $\{f_n\}$ is normal, then there is some subsequence $\{f_{n_k}\}$ which converges uniformly on compact subsets of $\mathbb{\Omega}$. So, if I take $m_n=\frac{1}{n}, n\in\mathbb{N}$, would this work? This feels like I am trying to prove the "other" definition of normal by saying the aforementioned subsequences must converge uniformly on compact subsets of $\mathbb{\Omega}$, OR tend to $\infty$ uniformly... but I suppose that just because $m_n\rightarrow\infty$, does not necessarily imply that $\{f_n^{m_n}\}\rightarrow\infty$. Any help is greatly appreciated! Thank you.
Consider $\{z\}$ as the normal family on $\mathbb{C}$
Then for any $m_n\to \infty$, the family $\{z^{m_n}\}$ is easily seen not to be normal.