If a generating function for $f(n)$ is rational, $f(n)$ cannot be more than exponential. If I have a generating function for $f(n)$ defined by
$g(x)=\displaystyle\sum_{n=0}^{\infty}f(n)x^n=\dfrac{P(x)}{Q(x)}$,
where $P(x)$ and $Q(x)$ are polynomials and $Q(x)$ is not the zero function, how could I show that $f(n)$ is not more than exponential?
 A: *

*Since $Q(0)\ne0$, $|Q(z)|\geqslant a$ for every $z$ in a neighborhood of $0$, say for every $|z|\leqslant\varepsilon$, for some $a\gt0$ and $\varepsilon\gt0$. 

*Since $P$ is continuous, $|P(z)|\leqslant b$ for every $|z|\leqslant\varepsilon$, for some finite $b$. 

*For every $n$,
$$
f(n)=\frac1{2\mathrm i\pi}\oint\frac{P(z)}{Q(z)}\frac{\mathrm dz}{z^{n+1}},
$$
where the integral is over the circle of equation $|z|=\varepsilon$, whose length is $2\pi\varepsilon$. 

*In particular,
$$
|f(n)|\leqslant\frac1{2\pi}2\pi\varepsilon\frac{b}a\frac1{\varepsilon^{n+1}}=cK^n,
$$
with $c=b/a$ and $K=1/\varepsilon$.

A: If you look backward to your problem you see that $f(n)$ is nothing, but the nth derivative of your rational polynomial $\frac{P(x)}{Q(x)}$ divided by $ n! $. I refer you to my Ph.D thesis (section 6.2.1) which covers the topic of finding the nth derivative of any rational polynomial. In other words you get a closed form formula for the nth derivative which may help you to conclude you assertion:
https://docs.google.com/file/d/0B4FXAHVyGS9KMGRiNDMyNDctMmQ1NS00MDI3LTk2OWEtNzc3N2ZlNDVmYjJm/edit?hl=en_GB&pli=1
