1962 AMC 12 Problem 40

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.

Problem 40 in other years