Calculate $0,2\overline{91}$ using geometric series

Question

I need to calculate $0,2\overline{91}$ using geometric series.

Not sure if these steps are correct:

$$0,2\overline{91} = \frac{2}{10} + \sum_{n=1}^{\infty}\frac{91}{(1000)^n}$$

so: $$0,2\overline{91} = \frac{2}{10} + 91 \sum_{n=1}^{\infty}\frac{1}{(1000)^n} $$ $$0,2\overline{91} = \frac{2}{10} + 91 \bigg(\frac{1}{1- \frac{1}{1000}} -1\bigg) $$ $$0,2\overline{91} = \frac{2}{10} + 91 \bigg(\frac{1000}{999} - 1\bigg) $$ $$0,2\overline{91} = \frac{2}{10} + 91 \bigg(\frac{1000 - 999}{999}\bigg) $$ $$0,2\overline{91} = \frac{2}{10} + \frac{91}{999} $$

is this correct?

Answer 1

No, it is not correct. That $1000$ should be $100$. I think that it's better to see it this way: if your number is $x$, then\begin{align}10x&=2,\overline{91}\\&=2+\sum_{n=1}^\infty\frac{91}{100^n}\\&=2+\frac{91}{99}\end{align}and therefore$$x=\frac15+\frac{91}{990}.$$

Answer 2

No it is not correct. Another point of view is to write $$ x=0.0\overline{91} $$ then, by periodicity, $$ 1000x=91+10x,\qquad 990x=91, $$that is $$ 0.0\overline{91}=\frac{91}{990} $$and

$$ 0.\color{red}{2}\overline{91}=\color{red}{\frac{1}{5}}+\frac{91}{990}=\frac{289}{990}. $$