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1962 AHSME Problems/Problem 40

Problem

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:

$\textbf{(A)}\ \frac{1}9\qquad\textbf{(B)}\ \frac{10}{81}\qquad\textbf{(C)}\ \frac{1}8\qquad\textbf{(D)}\ \frac{17}{72}\qquad\textbf{(E)}\ \text{larger than any finite quantity}$

Solution

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 $\frac{\frac{\frac{1}{10}}{1-\frac{1}{10}}}{\frac{9}{10}}$ $=$ $\frac{\frac{\frac{1}{10}}{\frac{9}{10}}}{\frac{9}{10}}$ $=$ $\frac{\frac{1}{9}}{\frac{9}{10}}$ $=$ $\boxed{ (B) \frac{10}{81}}$

Solution 2

So.. we have the sum to be $\frac{1}{10}+\frac{2}{100}+\frac{3}{1000}$... Notice that this can be written as $\frac{1}{10}+\frac{0.2}{10}+\frac{0.03}{10}+\frac{0.004}{10}$. Now, it is trivial that the new fraction we seek is $\frac{1.234567891011......}{10}$

Testing the answer choices, we see that $\boxed{B}$ is the correct answer.

Solution 3

Let \[S = \frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + ...\] Then \[10S = 1 + \frac{2}{10} + \frac{3}{10^2} + \frac{4}{10^3} + ...\] Subtracting $1S$ from $10S$, we got: \begin{align*} 9S &= 1 + \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + ... \\ &= \frac{1}{1-\frac{1}{10}} = \frac{10}{9} \\ S &= \frac{10}{81} \end{align*} Therefore, the answer is $\boxed{(B) \frac{10}{81}}$. -nullptr07

Video Solution

Problem starts at 2:20 : https://www.youtube.com/watch?v=3PDZtddYQoM&t=5s

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