Answer
To prove: $\dfrac{p}{10^{s}-1}=0.\overline{p_{1}\dots p_{s}}$, where $p=p_{1}\dots p_{s}$ and $p_{1}, \dots ,p_{s}$ are digits for number $p$.
Proof:
Right hand side
$=0.\overline{p_{1}\dots p_{s}}$
$=0.p_{1}\dots p_{s} +0.\underbrace{0\dots 0}_\text{s zeroes}p_{1}\dots p_{s}+0.\underbrace{0\dots 0}_\text{s zeroes}\underbrace{0\dots 0}_\text{s zeroes}p_{1}\dots p_{s}+\cdots$
$=\dfrac{p_{1}\dots p_{s}}{10^{s}}+\dfrac{p_{1}\dots p_{s}}{10^{2s}}+\dfrac{p_{1}\dots p_{s}}{10^{3s}}+\cdots$
$=\dfrac{p}{10^{s}}+\dfrac{p}{10^{2s}}+\dfrac{p}{10^{3s}}+\cdots$
$=p\left(\dfrac{1}{10^{s}}+\dfrac{1}{10^{2s}}+\dfrac{1}{10^{3s}}+\cdots \right)$
$=p\cdot \dfrac{\frac{1}{10^{s}}}{1-\frac{1}{10^{s}}}$ [using the sum of infinite geometric series]
$=p\cdot \dfrac{1}{10^{s}-1}$
$=\dfrac{p}{10^{s}-1}$
$=$ Left hand side
Hence, proved.
Now, $r=\dfrac{2}{11}=\dfrac{18}{10^{2}-1}=0.\overline{18}=0.181818\dots .$
Work Step by Step
The results used are as follows:
(i) If a finite decimal expansion is $0.p_{1}\dots p_{s}$, then, $0.p_{1}\dots p_{s}=\dfrac{p_{1}\dots p_{s}}{10^{s}}$.
(ii) If we have $a +a\cdot r +a\cdot r^{2}+\cdots$ to be an infinite geometric series with common ratio, $r$ such that $-1
