What is that geometric series thing he was talking about?

What is that geometric series thing he was talking about? (I know the way that AoPS teaches, but I never thought of a way with geometric series.)

@professionalbronco sorry for getting to you so late!
here's the idea with a geometric series:
say we have some repeating decimal, let's use \(0.\overline{81}\).
If we break it down into its repeating parts, it might look something like this:
$$0.\overline{81}=0.810000+0.008100+0.000081+...$$ so on and so forth.
Each term has the 81 shifted over by two decimal places, so in other words, each term is \(\frac{1}{100}\) of the previous term.From here, there's two ways to tackle an infinite geometric series: one way is to work it out algebraically.
Let's call the infinite sum S, then:
\(S=\frac{81}{100}+\frac{81}{10000}+\frac{81}{1000000}+...\)
Since each term is \(\frac{1}{100}\) of the previous term, we can manipulate the equation by multiplying S by \(\frac{1}{100}\).\(\frac{S}{100}=\frac{81}{10000}+\frac{81}{1000000}+\frac{81}{100000000}+...\)
wait... so every term in \(\frac{S}{100}\) is going to match up with every term in \(S\), except for the initial \(\frac{81}{100}\). So, if we take \(S–\frac{S}{100}\), everything cancels out and we get\(S–\frac{S}{100} = \frac{81}{100} \implies \frac{99S}{100}=\frac{81}{100} \implies \boxed{S=\frac{81}{99}=\frac{9}{11}}\)
The other method is just plugging into the formula for a geometric series, where \(a\) is the first term and \(r\) is the common ratio: \(\frac{a}{1r}\)
This formula actually comes just from using the method in the first part, just with variables instead. Give it a shot!
Anyways, plugging \(a=\frac{81}{100}\) and \(r=\frac{1}{100}\) we get \(\frac{\frac{81}{100}}{1\frac{1}{100}}\implies \frac{\frac{81}{100}}{\frac{99}{100}}\implies \frac{81}{99}\) as with the first method.
Also, just using the patterns with repeating decimals, since \(0.\overline{81}\) repeats with a period of 2 (it repeats every 2 digits), we know its denominator is \(99\).

Thank you