The limiting sum of the infinite series, $\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots$ whose $n$th term is $\frac{n}{10^n}$ is: 
$\textbf{(A)}\ \frac{1}9\qquad\textbf{(B)}\ \frac{10}{81}\qquad\textbf{(C)}\ \frac{1}8\qquad\textbf{(D)}\ \frac{17}{72}\qquad\textbf{(E)}\ \text{larger than any finite quantity}$
The series can be written as the following:
$\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + ...$
$+ \frac{1}{10^2} + \frac{1}{10^3} + \frac{1}{10^4} + ...$
$+ \frac{1}{10^3} + \frac{1}{10^4} + \frac{1}{10^5} + ...$
and so on.
by using the formula for infinite geometric series $(\frac{a}{1-r})$,
We can get $\frac{\frac{1}{10}}{1-\frac{1}{10}}$ $+$ $\frac{\frac{1}{10^2}}{1-\frac{1}{10}}$ $+$ $\frac{\frac{1}{10^3}}{1-\frac{1}{10}}$ $+$ ...
Since they all have common denominators, we get $\frac{(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3})}{\frac{9}{10}}$.
Using the infinite series formula again, we get $\frac{\frac{\frac{1}{10}}{1-\frac{1}{10}}}{\frac{9}{10}}$ $=$ $\frac{\frac{\frac{1}{10}}{\frac{9}{10}}}{\frac{9}{10}}$ $=$ $\frac{\frac{1}{9}}{\frac{9}{10}}$ $=$ $\boxed{ (B) \frac{10}{81}}$