Question

The limiting sum of the infinite series $\frac1{10}+\frac2{10^2}+\frac3{10^3}+\cdots,$ whose $n$ th term is $\frac{n}{10^n},$ is:

$\dfrac19$

$\dfrac{10}{81}$

$\dfrac18$

$\dfrac{17}{72}$

larger than any finite quantity

Start from $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$

Differentiate and then multiply by $x$

For $|x|\lt1,$ $\sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}.$ Taking $x=\frac{1}{10}$ gives $\frac{\frac{1}{10}}{(\frac{9}{10})^2}=\frac{10}{81}.$

Thus, the correct answer is B.