10. (20 points) Given a positive integer $n(n \geqslant 3), a_{n}=$ $(\overline{1.00 \dot{0}})_{n}$ (i.e., $a_{n}$ equals the number represented in base $n$ as $\overline{1.00 \dot{2}}$). Try to find
$$
\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{1}{a_{3}}+\frac{1}{a_{3} a_{4}}+\cdots+\frac{1}{a_{3} a_{4} \cdots a_{n}}\right)
$$

the value.
10. According to the problem, we have
$$
\begin{array}{l}
a_{n}=\overline{1 . \dot{0} \dot{2}}=1+\frac{2}{n^{3}}+\frac{2}{n^{6}}+\cdots=1+\frac{\frac{2}{n^{3}}}{1-\frac{1}{n^{3}}} \\
=\frac{1+\frac{1}{n^{3}}}{1-\frac{1}{n^{3}}}=\frac{n^{3}+1}{n^{3}-1}=\frac{n+1}{n-1} \cdot \frac{n^{2}-n+1}{n^{2}+n+1} .
\end{array}
$$

Notice that
$$
n^{2}-n+1=(n-1)^{2}+(n-1)+1 \text {. }
$$

Then $a_{3} a_{4} \cdots a_{n}$
$$
\begin{array}{l}
=\prod_{k=3}^{n}\left[\frac{k+1}{k-1} \cdot \frac{(k-1)^{2}+(k-1)+1}{k^{2}+k+1}\right] \\
=\frac{n(n+1)}{2 \times 3} \cdot \frac{2^{2}+2+1}{n^{2}+n+1} \\
=\frac{7}{6} \cdot \frac{n(n+1)}{n(n+1)+1} .
\end{array}
$$

Therefore, $\frac{1}{a_{3} a_{4} \cdots a_{n}}=\frac{6}{7}\left[1+\frac{1}{n(n+1)}\right]$
$$
=\frac{6}{7}\left(1+\frac{1}{n}-\frac{1}{n+1}\right) \text {. }
$$

Thus, we have
$$
\begin{array}{l}
\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{1}{a_{3}}+\frac{1}{a_{3} a_{4}}+\cdots+\frac{1}{a_{3} a_{4} \cdots a_{n}}\right) \\
=\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \frac{6}{7} \sum_{k=3}^{n}\left(1+\frac{1}{k}-\frac{1}{k+1}\right) \\
=\lim _{n \rightarrow \infty} \frac{6}{7} \cdot \frac{1}{n}\left[(n-2)+\frac{1}{3}-\frac{1}{n+1}\right] \\
=\frac{6}{7} .
\end{array}
$$