$10 \cdot 12$ The sequence $\left\{\frac{100^{n}}{n!}\right\}(n=1,2,3, \cdots)$ is
(A) an increasing sequence.
(B) a decreasing sequence.
(C) a sequence that has both increasing and decreasing terms after the 1993rd term.
(D) a sequence for which there exists a term after which it is decreasing.
(High School Mathematics Competition in Taiyuan City, Shanxi Province, China, 1993)
[Solution] Given $a_{n}=\frac{100^{n}}{n!}, a_{n+1}=\frac{100^{n+1}}{(n+1)!}$,
$\therefore\left\{a_{n}\right\}$ is decreasing $\Longleftrightarrow \frac{a_{n+1}}{a_{n}}<1 \Longleftrightarrow \frac{100}{n+1}<1 \Longleftrightarrow n>99$. That is, from the 99th term onwards, $\left\{a_{n}\right\}$ is decreasing.
Therefore, the answer is $(D)$.