Question

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

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

Answer 1

The correct option is B $\frac{10}{81}$

Solution:- (B) $\frac{10}{81}$

Let $S=\frac{1}{10}+\frac{2}{{10}^2}+\frac{3}{{10}^3}+.............\left(1\right)$

Therefore,

$\frac{S}{10}=\frac{1}{{10}^2}+\frac{2}{{10}^3}+.............\left(2\right)$

Subtracting ${eq}^n\left(2\right)$ from $\left(1\right)$

$S-\frac{S}{10}=(\frac{1}{10}+\frac{2}{{10}^2}+\frac{3}{{10}^3}+........)-(\frac{1}{{10}^2}+\frac{2}{{10}^3}+\frac{3}{{10}^4}+........)$

$\Rightarrow \frac{9S}{10}=\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+.......$

$\Rightarrow 9S=1+\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+.......$

As the R.H.S. is a series in G.P. of infinite terms and sum of infinite series in G.P. is given by-

$S_{\infty }=\frac{a}{1-r}$

$\therefore 9S=\frac{1}{1-\left(\frac{1}{10}\right)}$

$\Rightarrow 9S=\frac{10}{9}$

$\Rightarrow S=\frac{10}{81}$

Hence the limiting sum of the given infinite series is $\frac{10}{81}$.