Question

The limiting sum of the infinite series, $\frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\dots$. whose $n^{\text{th }}$ term is $\frac{n}{10^n}$ is equal to

• A. $\frac{1}{9}$
• B. $\frac{10}{81}$
• C. $\frac{1}{8}$
• D. $\frac{17}{72}$

Answer 1

Answer: (B)

Let $S=\frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\dots \dots \dots ..$(1)
So, $\frac{S}{10}=0+\frac{1}{10^2}+\frac{2}{10^3}+\dots \dots \dots$(2)
$\therefore$ Subtracting (2) from (1), we get
$\frac{9S}{10}=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\dots \dots \dots ...=\frac{1/10}{1-\frac{1}{10}}=\frac{1}{9}$
$\Rightarrow S=\frac{1}{9}\left(\frac{10}{9}\right)=\frac{10}{81}$