Compactness of $\ A= \{f: f$ is power series with infinite radius of convergence and is bounded by $1\}$ Consider the set $\ K=C[0,1]$, the set of continuous functions on $\ [0,1]$, with the supremum norm. Let 
$\ A= \{f: f$ is power series with infinite radius of convergence and is bounded by $1\}$.
I am asked to prove or disprove whether A is compact or not.
I tried noting a counterexample to show that the set isn't closed.
$$ \sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}=\sin(x) $$
This is a power series which converges to $\sin(x)$ which isn't in the set. Does this work?

Let $f_n(x) = x^n, n=1,2,\dots $ Then each $f_n\in A.$ If $A$ were compact, then some subsequence $f_{n_k}$ would converge uniformly to some $f\in A.$ This would imply $f_{n_k}\to f$ pointwise on $[0,1].$ But the full sequence $f_n,$ hence the subsequence $f_{n_k},$ converges pointwise to the function that equals $0$ on $[0,1)$ and equals $1$ at $1.$ That function isn't even in $K,$ much less $A,$ contradiction. Thus $A$ is not compact. 
