5. The product of the first $n$ positive integers starting from 1 is denoted as $n!=1 \times 2 \times \cdots \times n$, for example, $3!=1 \times 2 \times 3=6$. Then $\sum_{n=2}^{8} \frac{n-1}{n!}=(\quad$.
( A) $\frac{719}{720}$
(B) $\frac{5039}{5040}$
(C) $\frac{40319}{40320}$
(D) $\frac{40321}{40320}$
5. C.

Notice,
$$
\begin{array}{l}
\frac{n-1}{n!}=\frac{n}{n!}-\frac{1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!} . \\
\text { Hence } \sum_{n=2}^{8} \frac{n-1}{n!}=\sum_{n=2}^{8}\left[\frac{1}{(n-1)!}-\frac{1}{n!}\right] \\
=1-\frac{1}{8!}=1-\frac{1}{40320}=\frac{40319}{40 \cdot 320} .
\end{array}
$$