10.80 Given the infinite series $\frac{1}{10}+\frac{2}{10^{2}}+\frac{3}{10^{3}}+\cdots$, where the $n$-th term is $\frac{n}{10^{n}}$, then its limit sum is
(A) $\frac{1}{9}$.
(B) $\frac{10}{81}$.
(C) $\frac{1}{8}$.
(D) $\frac{17}{72}$.
(E) greater than any finite number.
(13th American High School Mathematics Examination, 1962)
[Solution] Let $S_{n}=\frac{1}{10}+\frac{2}{10^{2}}+\frac{3}{10^{3}}+\cdots+\frac{n}{10^{n}}$.
Consider $S_{n}-\frac{S_{n}}{10}=\sum_{k=1}^{n} \frac{k}{10^{k}}-\sum_{k=1}^{n} \frac{k}{10^{k+1}}=\sum_{k=1}^{n} \frac{k}{10^{k}}-\sum_{k=2}^{n+1} \frac{k-1}{10^{k}}$
$$
\begin{array}{l}
=\left(\frac{1}{10}+\sum_{k=2}^{n} \frac{k}{10^{k}}\right)-\left(\sum_{k=2}^{n} \frac{k-1}{10^{k}}+\frac{n}{10^{n+1}}\right) \\
=\frac{1}{10}+\sum_{k=2}^{n}\left(\frac{k}{10^{k}}-\frac{k-1}{10^{k}}\right)-\frac{n}{10^{n+1}} \\
=\frac{1}{10}+\sum_{k=2}^{n} \frac{1}{10^{k}}-\frac{n}{10^{n+1}}=\sum_{k=1}^{n} \frac{1}{10^{k}}-\frac{n}{10^{n+1}},
\end{array}
$$

That is, $\frac{9}{10} S_{n}=\sum_{k=1}^{n} \frac{1}{10^{k}}-\frac{n}{10^{n+1}}$ or $S_{n}=\frac{10}{9} \sum_{k=1}^{n} \frac{1}{10^{k}}-\frac{n}{9 \cdot 10^{n}}$.
And $\lim _{n \rightarrow \infty} S_{n}=\frac{10}{9} \cdot \frac{\frac{1}{10}}{1-\frac{1}{10}}-0=\frac{10}{9} \cdot \frac{1}{9}=\frac{10}{81}$.
Therefore, the answer is $(B)$.