Q:

When does an analytic function grow faster than a polynomial? 

Suppose $f$ is an analytic function with power series expansion $f(z)=\sum_{n=0}^{\infty} a_nz^n$, and $p = \sum_{n=0}^{d}b_nz^n$ is a polynomial.  If $f$ is a polynomial of degree larger than $d$, then $|f|$ grows faster than $|p|$, but the situation is not so clear when the expansion of $f$ has infinitely many nonzero coefficients.  I would expect the growth of the function $f$ then to be faster than that of $p$, as with the function $e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!}$. However the function $\frac{1}{1-z} = \sum_{n=0}^{\infty}z^n$ also has infinitely many nonzero coefficients and grows slower than any polynomial (as $|z|\to\infty$).  I realize this is related to the failure of the power series to converge outside a disk of radius $1$. Also, $log(z)$ grows slower than any polynomial, but any power series representation cannot converge on an infinite radius (The function itself cannot be well-defined everywhere in the complex plane simultaneously). 
Under what conditions can we say that a power series with infinitely many nonzero coefficients represents a function that grows faster than any polynomial?  Is this true for any power series with infinite radius of convergence?  Are there such power series which grow at the rate $z^\alpha$, for any $\alpha\in(0,\infty)$? 
I have in mind the case where $f$ is complex-analytic, but I would also be interested to hear about the case where $f$ is real-analytic, if the cases differ. 

A:

Suppose a non-polynomial function $f$ has a power series
$$
f(z) = \sum_{n=0}^{\infty} a_n z^n
$$
which converges on all of $\mathbb{C}$.  Then for each integer $n \geq 0$ and all $r > 0$ we have
$$
\begin{align}
|a_n| &= \left|\frac{f^{(n)}(0)}{n!}\right| \\
&= \left| \frac{1}{2\pi} \int_{|z| = r} \frac{f(z)}{z^{n+1}}\,dz \right| \\
&\leq \frac{M(r)}{r^n}
\end{align}
$$
by Cauchy's integral theorem, where $M(r) = \max\limits_{|z| = r} |f(z)|$.  Since there are infinitely many nonzero coefficients $a_n$ we may conclude from this that $M(r)$ grows faster than any polynomial as $r \to \infty$.

