Representing a Decimal as a Fraction - 2 Methods So I am trying to represent the number 0.71717171 · · · as a fraction and have managed to do it using algebra. I was told I was supposed to solve it using a geometric sum.
Could someone guide me through the steps of solving such a question with geometric sums?
My method:
Let $x = 0.\overline{71}$
Then $100x = 71.\overline{71}$
Consequently, $$100x - x = 71.\overline{71} - 0.\overline{71}$$
$$99x = 71$$
$$x=\frac{71}{99}$$
Any help would be greatly appreciated. Thank you.
 A: $$0.71717171\cdots=0.71[1+.01+.0001+\cdots]$$
and
$$1+.01+.0001+\cdots=1+(0.01)+(0.01)^2+\cdots=\dfrac1{1-0.01}$$ as the common ratio$(0.01)$ lies in $(-1,1)$
A: Using geometric sums:

$$x=\dfrac{7}{10}+\dfrac{1}{10^2}+\dfrac{7}{10^3}+\dfrac{1}{10^4}+\dfrac{7}{10^5}+\ldots$$
  $$x= \left( \dfrac{7}{10}+\dfrac{7}{10^3}+\dfrac{7}{10^5}+\ldots \right)+\left( \dfrac{1}{10^2}+\dfrac{1}{10^4}+\dfrac{1}{10^6}+\ldots \right) $$
  Here you have a decreasing geometric  with ratio $r<1$, so:
  $$x=\left( \dfrac{\dfrac{7}{10}}{1-\dfrac{1}{10^2}} \right)+ \left( \dfrac{\dfrac{1}{10^2}}{1-\dfrac{1}{10^2}} \right)$$
  $$x=\frac{71}{99}$$

A: I think you have done fine.
One way of summing a geometric progression is to take:
$$S=\sum_{r=0}^nax^r$$
$$S-xS=a(1-x^{n+1})$$
$$S=a\cdot\frac{1-x^{n+1}}{1-x}$$
And if $-1\lt x\lt 1$, in the limit as $n\to \infty$ $$S=\frac a{1-x}$$
The process you have used echoes this process for the general case. In fact it relies on a repeating decimal being a geometric progression. If you are asked in a question to use the fact that this is a geometric progression, you simply need to make this explicit.
As you can see from the other answers, your method is simpler and less prone to error, and I have always preferred this direct method for use with repeating decimals, because I find it easier to see what is going on (especially when there is a non-repeating bit).
