The Decimal Series

Let $S = \displaystyle \sum_{n=0}^{\infty} 10^{-n} = 1 + \dfrac{1}{10} + \dfrac{1}{100} + \dfrac{1}{1000} + \dfrac{1}{10000} + ...... = 1.1111.......$ .

Then $10S = 11.1111......... = 10 + 1.1111......... = 10 + S \Longrightarrow 10S - S = 10 \Longrightarrow S = \boxed{\dfrac{10}{9}}$ .

simple and easy to solve.

just look at the infinite series.

we know that, summation of infinite series is = $\frac{a}{1-r}$

here , a(the 1st term)=1 and r(the ratio= $\frac{1}{10}$

so, the infinite summation , S= $\frac{1}{1-.1}$ = $\frac{10}{9}$ .............[ $\frac{1}{10}=.1$ ]

$\begin{aligned} S & = \displaystyle \sum_{n \ = \ 0}^{\infty} 10^{- n} \\ & = \sum_{n \ = \ 0}^{\infty} \frac {1}{10}^n \\ & = \frac {1}{1 - \frac {1}{10}} \\ & = \frac {10}{9} \end{aligned}$

BREAK THE SUM. YOU WILL GET A GEOMETRIC PROGRESSION. SOLVE USING ITS SUM.