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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.

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    $\begingroup$ You made a slight mistake in your comment on $\frac{1}{1-z}$. It is only equal to $\sum_{n=0}^{\infty}z^n$ if $|z| < 1$. It is not analytic outside of $|z| < 1$ (it is given by a Laurent expansion though) so it doesn't quite fit into your paradigm as stated in the title. You also have to be careful about what you mean when you say grows at infinity with holomorphic functions. $\exp(z^2)$ is holomorphic but along the $y$ axis, it is equal to $\exp(-y^2)$ which is a decaying function. I think you need to better formulate your idea. $\endgroup$ – Cameron Williams Jul 24 '13 at 0:12
  • $\begingroup$ If $f$ is an entire function, then either $f$ is a polynomial or $\frac{\lvert f(z)\rvert}{(1+\lvert z\rvert)^k}$ is unbounded for all $k$. $\endgroup$ – Daniel Fischer Jul 24 '13 at 0:28
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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 $r \to \infty$.

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    $\begingroup$ Thanks. I like this - using the ML-inequality in reverse. This was the type of argument I was looking for. I suppose in general it would be very difficult to say something about the growth rate when the series has a finite radius of convergence, even assuming the function can be analytically continued almost everywhere? $\endgroup$ – PrimeRibeyeDeal Jul 24 '13 at 1:04
  • $\begingroup$ If $f$ is not a rational function and is analytic on $\mathbb{C}$ except at finitely many singular points $z_k$ then it will still grow faster than any polynomial. To see this, let $P_k(z)$ be the principal part of the Laurent series of $f$ at the point $z_k$. Then $P_k(z) \to 0$ as $|z| \to \infty$ for each $k$ and $f - \sum_k P_k$ is entire and non-polynomial. We can then apply the argument in my answer to this function and deduce that it's also true for $f$. $\endgroup$ – Antonio Vargas Jul 24 '13 at 5:42
  • $\begingroup$ I'm not sure what the idea would be to address the behavior of functions with infinitely many singularities (like the gamma function). $\endgroup$ – Antonio Vargas Jul 24 '13 at 5:43

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