19. Let
$$
F(x)=\frac{1}{\left(2-x-x^{5}\right)^{2011}},
$$
and note that $F$ may be expanded as a power series so that $F(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$. Find an ordered pair of positive real numbers $(c, d)$ such that $\lim _{n \rightarrow \infty} \frac{a_{n}}{n^{d}}=c$.
Answer: $\left(\frac{1}{6^{2011} 2010!}, 2010\right)$ First notice that all the roots of $2-x-x^{5}$ that are not 1 lie strictly outside the unit circle. As such, we may write $2-x-x^{5}$ as $2(1-x)\left(1-r_{1} x\right)\left(1-r_{2} x\right)\left(1-r_{3} x\right)\left(1-r_{4} x\right)$ where $\left|r_{i}\right|<1$, and let $\frac{1}{\left(2-x-x^{5}\right)}=\frac{b_{0}}{(1-x)}+\frac{b_{1}}{\left(1-r_{1} x\right)}+\ldots+\frac{b_{4}}{\left(1-r_{4} x\right)}$. We calculate $b_{0}$ as $\lim _{x \rightarrow 1} \frac{(1-x)}{\left(2-x-x^{5}\right)}=$ $\lim _{x \rightarrow 1} \frac{(-1)}{\left(-1-5 x^{4}\right)}=\frac{1}{6}$.
Now raise the equation above to the 2011th power.
$$
\frac{1}{\left(2-x-x^{5}\right)^{2011}}=\left(\frac{1 / 6}{(1-x)}+\frac{b_{1}}{\left(1-r_{1} x\right)}+\ldots+\frac{b_{4}}{\left(1-r_{4} x\right)}\right)^{2011}
$$

Expand the right hand side using multinomial expansion and then apply partial fractions. The result will be a sum of the terms $(1-x)^{-k}$ and $\left(1-r_{i} x\right)^{-k}$, where $k \leq 2011$.
Since $\left|r_{i}\right|<1$, the power series of $\left(1-r_{i} x\right)^{-k}$ will have exponentially decaying coefficients, so we only need to consider the $(1-x)^{-k}$ terms. The coefficient of $x^{n}$ in the power series of $(1-x)^{-k}$ is $\binom{n+k-1}{k-1}$, which is a $(k-1)$ th degree polynomial in variable $n$. So when we sum up all coefficients, only the power series of $(1-x)^{-2011}$ will have impact on the leading term $n^{2010}$.
The coefficient of the $(1-x)^{-2011}$ term in the multinomial expansion is $\left(\frac{1}{6}\right)^{2011}$. The coefficient of the $x^{n}$ term in the power series of $(1-x)^{-2011}$ is $\binom{n+2010}{2010}=\frac{1}{2010!} n^{2010}+\ldots$. Therefore, $(c, d)=$ $\left(\frac{1}{6^{2011} 2010!}, 2010\right)$.