9. 44 Let $k$ be a natural number. Determine for which value of $k$, $A_{k}=\frac{19^{k}+66^{k}}{k!}$ attains its maximum value.
[Solution] When $1 \leqslant k \leqslant 18$, we have
$$
\begin{aligned}
A_{k+1}-A_{k} & =\frac{19^{k+1}}{(k+1)!}+\frac{66^{k+1}}{(k+1)!}-\frac{19^{k}}{k!}-\frac{66^{k}}{k!} \\
& =\frac{19^{k}}{k!}\left(\frac{19}{k+1}-1\right)+\frac{66^{k}}{k!}\left(\frac{66}{k+1}-1\right) \\
& =\frac{19^{k}}{k!} \cdot \frac{18-k}{k+1}+\frac{66^{k}}{k!} \cdot \frac{65-k}{k+1},
\end{aligned}
$$

so $A_{k+1}>A_{k}$. The following discussion is for $k \geqslant 19$. Since
$$
\begin{aligned}
(k+1)!\left(A_{k+1}-A_{k}\right. & =19^{k+1}+66^{k+1}-(k+1)\left(19^{k}+66^{k}\right) \\
& =66^{k}(66-k-1)-19^{k}(k+1-19),
\end{aligned}
$$

it is clear that $\quad A_{k+1}-A_{k}=\left\{\begin{array}{l}>0,19 \leqslant k \leqslant 64 ; \\ <0, k \geqslant 65 .\end{array}\right.$
From this, we know that when $k=65$, $A_{k}$ reaches its maximum value.